The explicit-control evaluator of 5.4 is a register machine whose controller interprets Scheme programs. In this section we will see how to run Scheme programs on a register machine whose controller is not a Scheme interpreter.
The explicit-control evaluator machine is universal—it can carry out any computational process that can be described in Scheme. The evaluator’s controller orchestrates the use of its data paths to perform the desired computation. Thus, the evaluator’s data paths are universal: They are sufficient to perform any computation we desire, given an appropriate controller.318
Commercial general-purpose computers are register machines organized around a collection of registers and operations that constitute an efficient and convenient universal set of data paths. The controller for a general-purpose machine is an interpreter for a register-machine language like the one we have been using. This language is called the native language of the machine, or simply machine language. Programs written in machine language are sequences of instructions that use the machine’s data paths. For example, the explicit-control evaluator’s instruction sequence can be thought of as a machine-language program for a general-purpose computer rather than as the controller for a specialized interpreter machine.
There are two common strategies for bridging the gap between higher-level languages and register-machine languages. The explicit-control evaluator illustrates the strategy of interpretation. An interpreter written in the native language of a machine configures the machine to execute programs written in a language (called the source language) that may differ from the native language of the machine performing the evaluation. The primitive procedures of the source language are implemented as a library of subroutines written in the native language of the given machine. A program to be interpreted (called the source program) is represented as a data structure. The interpreter traverses this data structure, analyzing the source program. As it does so, it simulates the intended behavior of the source program by calling appropriate primitive subroutines from the library.
In this section, we explore the alternative strategy of compilation. A compiler for a given source language and machine translates a source program into an equivalent program (called the object program) written in the machine’s native language. The compiler that we implement in this section translates programs written in Scheme into sequences of instructions to be executed using the explicit-control evaluator machine’s data paths.319
Compared with interpretation, compilation can provide a great increase in the efficiency of program execution, as we will explain below in the overview of the compiler. On the other hand, an interpreter provides a more powerful environment for interactive program development and debugging, because the source program being executed is available at run time to be examined and modified. In addition, because the entire library of primitives is present, new programs can be constructed and added to the system during debugging.
In view of the complementary advantages of compilation and interpretation, modern program-development environments pursue a mixed strategy. Lisp interpreters are generally organized so that interpreted procedures and compiled procedures can call each other. This enables a programmer to compile those parts of a program that are assumed to be debugged, thus gaining the efficiency advantage of compilation, while retaining the interpretive mode of execution for those parts of the program that are in the flux of interactive development and debugging. In 5.5.7, after we have implemented the compiler, we will show how to interface it with our interpreter to produce an integrated interpreter-compiler development system.
Our compiler is much like our interpreter, both in its structure and in the
function it performs. Accordingly, the mechanisms used by the compiler for
analyzing expressions will be similar to those used by the interpreter.
Moreover, to make it easy to interface compiled and interpreted code, we will
design the compiler to generate code that obeys the same conventions of
register usage as the interpreter: The environment will be kept in the
env
register, argument lists will be accumulated in argl
, a
procedure to be applied will be in proc
, procedures will return their
answers in val
, and the location to which a procedure should return will
be kept in continue
. In general, the compiler translates a source
program into an object program that performs essentially the same register
operations as would the interpreter in evaluating the same source program.
This description suggests a strategy for implementing a rudimentary compiler:
We traverse the expression in the same way the interpreter does. When we
encounter a register instruction that the interpreter would perform in
evaluating the expression, we do not execute the instruction but instead
accumulate it into a sequence. The resulting sequence of instructions will be
the object code. Observe the efficiency advantage of compilation over
interpretation. Each time the interpreter evaluates an expression—for
example, (f 84 96)
—it performs the work of classifying the expression
(discovering that this is a procedure application) and testing for the end of
the operand list (discovering that there are two operands). With a compiler,
the expression is analyzed only once, when the instruction sequence is
generated at compile time. The object code produced by the compiler contains
only the instructions that evaluate the operator and the two operands, assemble
the argument list, and apply the procedure (in proc
) to the arguments
(in argl
).
This is the same kind of optimization we implemented in the analyzing evaluator of 4.1.7. But there are further opportunities to gain efficiency in compiled code. As the interpreter runs, it follows a process that must be applicable to any expression in the language. In contrast, a given segment of compiled code is meant to execute some particular expression. This can make a big difference, for example in the use of the stack to save registers. When the interpreter evaluates an expression, it must be prepared for any contingency. Before evaluating a subexpression, the interpreter saves all registers that will be needed later, because the subexpression might require an arbitrary evaluation. A compiler, on the other hand, can exploit the structure of the particular expression it is processing to generate code that avoids unnecessary stack operations.
As a case in point, consider the combination (f 84 96)
. Before the
interpreter evaluates the operator of the combination, it prepares for this
evaluation by saving the registers containing the operands and the environment,
whose values will be needed later. The interpreter then evaluates the operator
to obtain the result in val
, restores the saved registers, and finally
moves the result from val
to proc
. However, in the particular
expression we are dealing with, the operator is the symbol f
, whose
evaluation is accomplished by the machine operation
lookup-variable-value
, which does not alter any registers. The compiler
that we implement in this section will take advantage of this fact and generate
code that evaluates the operator using the instruction
(assign proc (op lookup-variable-value) (const f) (reg env))
This code not only avoids the unnecessary saves and restores but also assigns
the value of the lookup directly to proc
, whereas the interpreter would
obtain the result in val
and then move this to proc
.
A compiler can also optimize access to the environment. Having analyzed the
code, the compiler can in many cases know in which frame a particular variable
will be located and access that frame directly, rather than performing the
lookup-variable-value
search. We will discuss how to implement such
variable access in 5.5.6. Until then, however, we will focus on
the kind of register and stack optimizations described above. There are many
other optimizations that can be performed by a compiler, such as coding
primitive operations “in line” instead of using a general apply
mechanism (see Exercise 5.38); but we will not emphasize these here. Our
main goal in this section is to illustrate the compilation process in a
simplified (but still interesting) context.
In 4.1.7 we modified our original metacircular interpreter to separate analysis from execution. We analyzed each expression to produce an execution procedure that took an environment as argument and performed the required operations. In our compiler, we will do essentially the same analysis. Instead of producing execution procedures, however, we will generate sequences of instructions to be run by our register machine.
The procedure compile
is the top-level dispatch in the compiler. It
corresponds to the eval
procedure of 4.1.1, the
analyze
procedure of 4.1.7, and the eval-dispatch
entry point of the explicit-control-evaluator in 5.4.1. The
compiler, like the interpreters, uses the expression-syntax procedures defined
in 4.1.2.320 Compile
performs a
case analysis on the syntactic type of the expression to be compiled. For each
type of expression, it dispatches to a specialized
code generator:
(define (compile exp target linkage) (cond ((self-evaluating? exp) (compile-self-evaluating exp target linkage)) ((quoted? exp) (compile-quoted exp target linkage)) ((variable? exp) (compile-variable exp target linkage)) ((assignment? exp) (compile-assignment exp target linkage)) ((definition? exp) (compile-definition exp target linkage)) ((if? exp) (compile-if exp target linkage)) ((lambda? exp) (compile-lambda exp target linkage)) ((begin? exp) (compile-sequence (begin-actions exp) target linkage)) ((cond? exp) (compile (cond->if exp) target linkage)) ((application? exp) (compile-application exp target linkage)) (else (error "Unknown expression type: COMPILE" exp))))
Compile
and the code generators that it calls take two arguments in
addition to the expression to compile. There is a
target, which
specifies the register in which the compiled code is to return the value of the
expression. There is also a
linkage descriptor, which describes how
the code resulting from the compilation of the expression should proceed when
it has finished its execution. The linkage descriptor can require that the
code do one of the following three things:
next
),
return
), or
For example, compiling the expression 5
(which is self-evaluating) with
a target of the val
register and a linkage of next
should produce
the instruction
(assign val (const 5))
Compiling the same expression with a linkage of return
should produce
the instructions
(assign val (const 5)) (goto (reg continue))
In the first case, execution will continue with the next instruction in the
sequence. In the second case, we will return from a procedure call. In both
cases, the value of the expression will be placed into the target val
register.
Each code generator returns an instruction sequence containing the object code it has generated for the expression. Code generation for a compound expression is accomplished by combining the output from simpler code generators for component expressions, just as evaluation of a compound expression is accomplished by evaluating the component expressions.
The simplest method for combining instruction sequences is a procedure called
append-instruction-sequences
. It takes as arguments any number of
instruction sequences that are to be executed sequentially; it appends them and
returns the combined sequence. That is, if
and are sequences of instructions, then
evaluating
(append-instruction-sequences ⟨seq₁⟩ ⟨seq₂⟩)
produces the sequence
⟨seq₁⟩ ⟨seq₂⟩
Whenever registers might need to be saved, the compiler’s code generators use
preserving
, which is a more subtle method for combining instruction
sequences. Preserving
takes three arguments: a set of registers and two
instruction sequences that are to be executed sequentially. It appends the
sequences in such a way that the contents of each register in the set is
preserved over the execution of the first sequence, if this is needed for the
execution of the second sequence. That is, if the first sequence modifies the
register and the second sequence actually needs the register’s original
contents, then preserving
wraps a save
and a restore
of
the register around the first sequence before appending the sequences.
Otherwise, preserving
simply returns the appended instruction sequences.
Thus, for example,
(preserving (list ⟨reg₁⟩ ⟨reg₂⟩) ⟨seg₁⟩ ⟨seg₂⟩)
produces one of the following four sequences of instructions, depending on how
and
use and :
By using preserving
to combine instruction sequences the compiler avoids
unnecessary stack operations. This also isolates the details of whether or not
to generate save
and restore
instructions within the
preserving
procedure, separating them from the concerns that arise in
writing each of the individual code generators. In fact no save
or
restore
instructions are explicitly produced by the code generators.
In principle, we could represent an instruction sequence simply as a list of
instructions. Append-instruction-sequences
could then combine
instruction sequences by performing an ordinary list append
. However,
preserving
would then be a complex operation, because it would have to
analyze each instruction sequence to determine how the sequence uses its
registers. Preserving
would be inefficient as well as complex, because
it would have to analyze each of its instruction sequence arguments, even
though these sequences might themselves have been constructed by calls to
preserving
, in which case their parts would have already been analyzed.
To avoid such repetitious analysis we will associate with each instruction
sequence some information about its register use. When we construct a basic
instruction sequence we will provide this information explicitly, and the
procedures that combine instruction sequences will derive register-use
information for the combined sequence from the information associated with the
component sequences.
An instruction sequence will contain three pieces of information:
We will represent an instruction sequence as a list of its three parts. The constructor for instruction sequences is thus
(define (make-instruction-sequence needs modifies statements) (list needs modifies statements))
For example, the two-instruction sequence that looks up the value of the
variable x
in the current environment, assigns the result to val
,
and then returns, requires registers env
and continue
to have
been initialized, and modifies register val
. This sequence would
therefore be constructed as
(make-instruction-sequence '(env continue) '(val) '((assign val (op lookup-variable-value) (const x) (reg env)) (goto (reg continue))))
We sometimes need to construct an instruction sequence with no statements:
(define (empty-instruction-sequence) (make-instruction-sequence '() '() '()))
The procedures for combining instruction sequences are shown in 5.5.4.
Exercise 5.31: In evaluating a procedure application, the explicit-control evaluator always saves and restores the
env
register around the evaluation of the operator, saves and restoresenv
around the evaluation of each operand (except the final one), saves and restoresargl
around the evaluation of each operand, and saves and restoresproc
around the evaluation of the operand sequence. For each of the following combinations, say which of thesesave
andrestore
operations are superfluous and thus could be eliminated by the compiler’spreserving
mechanism:(f 'x 'y) ((f) 'x 'y) (f (g 'x) y) (f (g 'x) 'y)
Exercise 5.32: Using the
preserving
mechanism, the compiler will avoid saving and restoringenv
around the evaluation of the operator of a combination in the case where the operator is a symbol. We could also build such optimizations into the evaluator. Indeed, the explicit-control evaluator of 5.4 already performs a similar optimization, by treating combinations with no operands as a special case.
- Extend the explicit-control evaluator to recognize as a separate class of expressions combinations whose operator is a symbol, and to take advantage of this fact in evaluating such expressions.
- Alyssa P. Hacker suggests that by extending the evaluator to recognize more and more special cases we could incorporate all the compiler’s optimizations, and that this would eliminate the advantage of compilation altogether. What do you think of this idea?
In this section and the next we implement the code generators to which the
compile
procedure dispatches.
In general, the output of each code generator will end with
instructions—generated by the procedure compile-linkage
—that
implement the required linkage. If the linkage is return
then we must
generate the instruction (goto (reg continue))
. This needs the
continue
register and does not modify any registers. If the linkage is
next
, then we needn’t include any additional instructions. Otherwise,
the linkage is a label, and we generate a goto
to that label, an
instruction that does not need or modify any registers.321
(define (compile-linkage linkage) (cond ((eq? linkage 'return) (make-instruction-sequence '(continue) '() '((goto (reg continue))))) ((eq? linkage 'next) (empty-instruction-sequence)) (else (make-instruction-sequence '() '() `((goto (label ,linkage)))))))
The linkage code is appended to an instruction sequence by preserving
the continue
register, since a return
linkage will require the
continue
register: If the given instruction sequence modifies
continue
and the linkage code needs it, continue
will be saved
and restored.
(define (end-with-linkage linkage instruction-sequence) (preserving '(continue) instruction-sequence (compile-linkage linkage)))
The code generators for self-evaluating expressions, quotations, and variables construct instruction sequences that assign the required value to the target register and then proceed as specified by the linkage descriptor.
(define (compile-self-evaluating exp target linkage) (end-with-linkage linkage (make-instruction-sequence '() (list target) `((assign ,target (const ,exp)))))) (define (compile-quoted exp target linkage) (end-with-linkage linkage (make-instruction-sequence '() (list target) `((assign ,target (const ,(text-of-quotation exp))))))) (define (compile-variable exp target linkage) (end-with-linkage linkage (make-instruction-sequence '(env) (list target) `((assign ,target (op lookup-variable-value) (const ,exp) (reg env))))))
All these assignment instructions modify the target register, and the one that
looks up a variable needs the env
register.
Assignments and definitions are handled much as they are in the interpreter.
We recursively generate code that computes the value to be assigned to the
variable, and append to it a two-instruction sequence that actually sets or
defines the variable and assigns the value of the whole expression (the symbol
ok
) to the target register. The recursive compilation has target
val
and linkage next
so that the code will put its result into
val
and continue with the code that is appended after it. The appending
is done preserving env
, since the environment is needed for setting or
defining the variable and the code for the variable value could be the
compilation of a complex expression that might modify the registers in
arbitrary ways.
(define (compile-assignment exp target linkage) (let ((var (assignment-variable exp)) (get-value-code (compile (assignment-value exp) 'val 'next))) (end-with-linkage linkage (preserving '(env) get-value-code (make-instruction-sequence '(env val) (list target) `((perform (op set-variable-value!) (const ,var) (reg val) (reg env)) (assign ,target (const ok)))))))) (define (compile-definition exp target linkage) (let ((var (definition-variable exp)) (get-value-code (compile (definition-value exp) 'val 'next))) (end-with-linkage linkage (preserving '(env) get-value-code (make-instruction-sequence '(env val) (list target) `((perform (op define-variable!) (const ,var) (reg val) (reg env)) (assign ,target (const ok))))))))
The appended two-instruction sequence requires env
and val
and
modifies the target. Note that although we preserve env
for this
sequence, we do not preserve val
, because the get-value-code
is
designed to explicitly place its result in val
for use by this sequence.
(In fact, if we did preserve val
, we would have a bug, because this
would cause the previous contents of val
to be restored right after the
get-value-code
is run.)
The code for an if
expression compiled with a given target and linkage
has the form
⟨compilation of predicate, targetval
, linkagenext
⟩ (test (op false?) (reg val)) (branch (label false-branch)) true-branch ⟨compilation of consequent with given target and given linkage orafter-if
⟩ false-branch ⟨compilation of alternative with given target and linkage⟩ after-if
To generate this code, we compile the predicate, consequent, and alternative,
and combine the resulting code with instructions to test the predicate result
and with newly generated labels to mark the true and false branches and the end
of the conditional.322 In this arrangement of code, we must branch around the true branch if the
test is false. The only slight complication is in how the linkage for the true
branch should be handled. If the linkage for the conditional is return
or a label, then the true and false branches will both use this same linkage.
If the linkage is next
, the true branch ends with a jump around the code
for the false branch to the label at the end of the conditional.
(define (compile-if exp target linkage) (let ((t-branch (make-label 'true-branch)) (f-branch (make-label 'false-branch)) (after-if (make-label 'after-if))) (let ((consequent-linkage (if (eq? linkage 'next) after-if linkage))) (let ((p-code (compile (if-predicate exp) 'val 'next)) (c-code (compile (if-consequent exp) target consequent-linkage)) (a-code (compile (if-alternative exp) target linkage))) (preserving '(env continue) p-code (append-instruction-sequences (make-instruction-sequence '(val) '() `((test (op false?) (reg val)) (branch (label ,f-branch)))) (parallel-instruction-sequences (append-instruction-sequences t-branch c-code) (append-instruction-sequences f-branch a-code)) after-if))))))
Env
is preserved around the predicate code because it could be needed by
the true and false branches, and continue
is preserved because it could
be needed by the linkage code in those branches. The code for the true and
false branches (which are not executed sequentially) is appended using a
special combiner parallel-instruction-sequences
described in
5.5.4.
Note that cond
is a derived expression, so all that the compiler needs
to do handle it is to apply the cond->if
transformer (from
4.1.2) and compile the resulting if
expression.
The compilation of sequences (from procedure bodies or explicit begin
expressions) parallels their evaluation. Each expression of the sequence is
compiled—the last expression with the linkage specified for the sequence, and
the other expressions with linkage next
(to execute the rest of the
sequence). The instruction sequences for the individual expressions are
appended to form a single instruction sequence, such that env
(needed
for the rest of the sequence) and continue
(possibly needed for the
linkage at the end of the sequence) are preserved.
(define (compile-sequence seq target linkage) (if (last-exp? seq) (compile (first-exp seq) target linkage) (preserving '(env continue) (compile (first-exp seq) target 'next) (compile-sequence (rest-exps seq) target linkage))))
lambda
expressionsLambda
expressions construct procedures. The object code for a
lambda
expression must have the form
⟨construct procedure object and assign it to target register⟩ ⟨linkage⟩
When we compile the lambda
expression, we also generate the code for the
procedure body. Although the body won’t be executed at the time of procedure
construction, it is convenient to insert it into the object code right after
the code for the lambda
. If the linkage for the lambda
expression is a label or return
, this is fine. But if the linkage is
next
, we will need to skip around the code for the procedure body by
using a linkage that jumps to a label that is inserted after the body. The
object code thus has the form
⟨construct procedure object
and assign it to target register⟩
⟨code for given linkage⟩ or
(goto (label after-lambda))
⟨compilation of procedure body⟩
after-lambda
Compile-lambda
generates the code for constructing the procedure object
followed by the code for the procedure body. The procedure object will be
constructed at run time by combining the current environment (the environment
at the point of definition) with the entry point to the compiled procedure body
(a newly generated label).323
(define (compile-lambda exp target linkage) (let ((proc-entry (make-label 'entry)) (after-lambda (make-label 'after-lambda))) (let ((lambda-linkage (if (eq? linkage 'next) after-lambda linkage))) (append-instruction-sequences (tack-on-instruction-sequence (end-with-linkage lambda-linkage (make-instruction-sequence '(env) (list target) `((assign ,target (op make-compiled-procedure) (label ,proc-entry) (reg env))))) (compile-lambda-body exp proc-entry)) after-lambda))))
Compile-lambda
uses the special combiner
tack-on-instruction-sequence
rather than
append-instruction-sequences
(5.5.4) to append the procedure body to the
lambda
expression code, because the body is not part of the sequence of
instructions that will be executed when the combined sequence is entered;
rather, it is in the sequence only because that was a convenient place to put
it.
Compile-lambda-body
constructs the code for the body of the procedure.
This code begins with a label for the entry point. Next come instructions that
will cause the run-time evaluation environment to switch to the correct
environment for evaluating the procedure body—namely, the definition
environment of the procedure, extended to include the bindings of the formal
parameters to the arguments with which the procedure is called. After this
comes the code for the sequence of expressions that makes up the procedure
body. The sequence is compiled with linkage return
and target
val
so that it will end by returning from the procedure with the
procedure result in val
.
(define (compile-lambda-body exp proc-entry) (let ((formals (lambda-parameters exp))) (append-instruction-sequences (make-instruction-sequence '(env proc argl) '(env) `(,proc-entry (assign env (op compiled-procedure-env) (reg proc)) (assign env (op extend-environment) (const ,formals) (reg argl) (reg env)))) (compile-sequence (lambda-body exp) 'val 'return))))
The essence of the compilation process is the compilation of procedure applications. The code for a combination compiled with a given target and linkage has the form
⟨compilation of operator, targetproc
, linkagenext
⟩ ⟨evaluate operands and construct argument list inargl
⟩ ⟨compilation of procedure call with given target and linkage⟩
The registers env
, proc
, and argl
may have to be saved and
restored during evaluation of the operator and operands. Note that this is the
only place in the compiler where a target other than val
is specified.
The required code is generated by compile-application
. This recursively
compiles the operator, to produce code that puts the procedure to be applied
into proc
, and compiles the operands, to produce code that evaluates the
individual operands of the application. The instruction sequences for the
operands are combined (by construct-arglist
) with code that constructs
the list of arguments in argl
, and the resulting argument-list code is
combined with the procedure code and the code that performs the procedure call
(produced by compile-procedure-call
). In appending the code sequences,
the env
register must be preserved around the evaluation of the operator
(since evaluating the operator might modify env
, which will be needed to
evaluate the operands), and the proc
register must be preserved around
the construction of the argument list (since evaluating the operands might
modify proc
, which will be needed for the actual procedure application).
Continue
must also be preserved throughout, since it is needed for the
linkage in the procedure call.
(define (compile-application exp target linkage) (let ((proc-code (compile (operator exp) 'proc 'next)) (operand-codes (map (lambda (operand) (compile operand 'val 'next)) (operands exp)))) (preserving '(env continue) proc-code (preserving '(proc continue) (construct-arglist operand-codes) (compile-procedure-call target linkage)))))
The code to construct the argument list will evaluate each operand into
val
and then cons
that value onto the argument list being
accumulated in argl
. Since we cons
the arguments onto
argl
in sequence, we must start with the last argument and end with the
first, so that the arguments will appear in order from first to last in the
resulting list. Rather than waste an instruction by initializing argl
to the empty list to set up for this sequence of evaluations, we make the first
code sequence construct the initial argl
. The general form of the
argument-list construction is thus as follows:
⟨compilation of last operand, targeted toval
⟩ (assign argl (op list) (reg val)) ⟨compilation of next operand, targeted toval
⟩ (assign argl (op cons) (reg val) (reg argl)) … ⟨compilation of first operand, targeted toval
⟩ (assign argl (op cons) (reg val) (reg argl))
Argl
must be preserved around each operand evaluation except the first
(so that arguments accumulated so far won’t be lost), and env
must be
preserved around each operand evaluation except the last (for use by subsequent
operand evaluations).
Compiling this argument code is a bit tricky, because of the special treatment
of the first operand to be evaluated and the need to preserve argl
and
env
in different places. The construct-arglist
procedure takes
as arguments the code that evaluates the individual operands. If there are no
operands at all, it simply emits the instruction
(assign argl (const ()))
Otherwise, construct-arglist
creates code that initializes argl
with the last argument, and appends code that evaluates the rest of the
arguments and adjoins them to argl
in succession. In order to process
the arguments from last to first, we must reverse the list of operand code
sequences from the order supplied by compile-application
.
(define (construct-arglist operand-codes) (let ((operand-codes (reverse operand-codes))) (if (null? operand-codes) (make-instruction-sequence '() '(argl) '((assign argl (const ())))) (let ((code-to-get-last-arg (append-instruction-sequences (car operand-codes) (make-instruction-sequence '(val) '(argl) '((assign argl (op list) (reg val))))))) (if (null? (cdr operand-codes)) code-to-get-last-arg (preserving '(env) code-to-get-last-arg (code-to-get-rest-args (cdr operand-codes)))))))) (define (code-to-get-rest-args operand-codes) (let ((code-for-next-arg (preserving '(argl) (car operand-codes) (make-instruction-sequence '(val argl) '(argl) '((assign argl (op cons) (reg val) (reg argl))))))) (if (null? (cdr operand-codes)) code-for-next-arg (preserving '(env) code-for-next-arg (code-to-get-rest-args (cdr operand-codes))))))
After evaluating the elements of a combination, the compiled code must apply
the procedure in proc
to the arguments in argl
. The code
performs essentially the same dispatch as the apply
procedure in the
metacircular evaluator of 4.1.1 or the apply-dispatch
entry point in the explicit-control evaluator of 5.4.1. It
checks whether the procedure to be applied is a primitive procedure or a
compiled procedure. For a primitive procedure, it uses
apply-primitive-procedure
; we will see shortly how it handles compiled
procedures. The procedure-application code has the following form:
(test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch)) compiled-branch ⟨code to apply compiled procedure with given target and appropriate linkage⟩ primitive-branch (assign ⟨target⟩ (op apply-primitive-procedure) (reg proc) (reg argl)) ⟨linkage⟩ after-call
Observe that the compiled branch must skip around the primitive branch.
Therefore, if the linkage for the original procedure call was next
, the
compound branch must use a linkage that jumps to a label that is inserted after
the primitive branch. (This is similar to the linkage used for the true branch
in compile-if
.)
(define (compile-procedure-call target linkage) (let ((primitive-branch (make-label 'primitive-branch)) (compiled-branch (make-label 'compiled-branch)) (after-call (make-label 'after-call))) (let ((compiled-linkage (if (eq? linkage 'next) after-call linkage))) (append-instruction-sequences (make-instruction-sequence '(proc) '() `((test (op primitive-procedure?) (reg proc)) (branch (label ,primitive-branch)))) (parallel-instruction-sequences (append-instruction-sequences compiled-branch (compile-proc-appl target compiled-linkage)) (append-instruction-sequences primitive-branch (end-with-linkage linkage (make-instruction-sequence '(proc argl) (list target) `((assign ,target (op apply-primitive-procedure) (reg proc) (reg argl))))))) after-call))))
The primitive and compound branches, like the true and false branches in
compile-if
, are appended using parallel-instruction-sequences
rather than the ordinary append-instruction-sequences
, because they will
not be executed sequentially.
The code that handles procedure application is the most subtle part of the
compiler, even though the instruction sequences it generates are very short. A
compiled procedure (as constructed by compile-lambda
) has an entry
point, which is a label that designates where the code for the procedure
starts. The code at this entry point computes a result in val
and
returns by executing the instruction (goto (reg continue))
. Thus, we
might expect the code for a compiled-procedure application (to be generated by
compile-proc-appl
) with a given target and linkage to look like this if
the linkage is a label
(assign continue
(label proc-return))
(assign val
(op compiled-procedure-entry)
(reg proc))
(goto (reg val))
proc-return
(assign ⟨target⟩
(reg val)) ; included if target is not val
(goto (label ⟨linkage⟩)) ; linkage code
or like this if the linkage is return
.
(save continue)
(assign continue
(label proc-return))
(assign val
(op compiled-procedure-entry)
(reg proc))
(goto (reg val))
proc-return
(assign ⟨target⟩
(reg val)) ; included if target is not val
(restore continue)
(goto (reg continue)) ; linkage code
This code sets up continue
so that the procedure will return to a label
proc-return
and jumps to the procedure’s entry point. The code at
proc-return
transfers the procedure’s result from val
to the
target register (if necessary) and then jumps to the location specified by the
linkage. (The linkage is always return
or a label, because
compile-procedure-call
replaces a next
linkage for the
compound-procedure branch by an after-call
label.)
In fact, if the target is not val
, that is exactly the code our compiler
will generate.324 Usually, however, the target is
val
(the only time the compiler specifies a different register is when
targeting the evaluation of an operator to proc
), so the procedure
result is put directly into the target register and there is no need to return
to a special location that copies it. Instead, we simplify the code by setting
up continue
so that the procedure will “return” directly to the place
specified by the caller’s linkage:
⟨set up continue
for linkage⟩
(assign val
(op compiled-procedure-entry)
(reg proc))
(goto (reg val))
If the linkage is a label, we set up continue
so that the procedure will
return to that label. (That is, the (goto (reg continue))
the procedure
ends with becomes equivalent to the (goto (label ⟨linkage⟩))
at
proc-return
above.)
(assign continue (label ⟨linkage⟩)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val))
If the linkage is return
, we don’t need to set up continue
at
all: It already holds the desired location. (That is, the (goto (reg
continue))
the procedure ends with goes directly to the place where the
(goto (reg continue))
at proc-return
would have gone.)
(assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val))
With this implementation of the return
linkage, the compiler generates
tail-recursive code. Calling a procedure as the final step in a procedure body
does a direct transfer, without saving any information on the stack.
Suppose instead that we had handled the case of a procedure call with a linkage
of return
and a target of val
as shown above for a non-val
target. This would destroy tail recursion. Our system would still give the
same value for any expression. But each time we called a procedure, we would
save continue
and return after the call to undo the (useless) save.
These extra saves would accumulate during a nest of procedure
calls.325
Compile-proc-appl
generates the above procedure-application code by
considering four cases, depending on whether the target for the call is
val
and whether the linkage is return
. Observe that the
instruction sequences are declared to modify all the registers, since executing
the procedure body can change the registers in arbitrary ways.326
Also note that the code sequence for the case with target val
and
linkage return
is declared to need continue
: Even though
continue
is not explicitly used in the two-instruction sequence, we must
be sure that continue
will have the correct value when we enter the
compiled procedure.
(define (compile-proc-appl target linkage) (cond ((and (eq? target 'val) (not (eq? linkage 'return))) (make-instruction-sequence '(proc) all-regs `((assign continue (label ,linkage)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val))))) ((and (not (eq? target 'val)) (not (eq? linkage 'return))) (let ((proc-return (make-label 'proc-return))) (make-instruction-sequence '(proc) all-regs `((assign continue (label ,proc-return)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) ,proc-return (assign ,target (reg val)) (goto (label ,linkage)))))) ((and (eq? target 'val) (eq? linkage 'return)) (make-instruction-sequence '(proc continue) all-regs '((assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val))))) ((and (not (eq? target 'val)) (eq? linkage 'return)) (error "return linkage, target not val: COMPILE" target))))
This section describes the details on how instruction sequences are represented and combined. Recall from 5.5.1 that an instruction sequence is represented as a list of the registers needed, the registers modified, and the actual instructions. We will also consider a label (symbol) to be a degenerate case of an instruction sequence, which doesn’t need or modify any registers. So to determine the registers needed and modified by instruction sequences we use the selectors
(define (registers-needed s) (if (symbol? s) '() (car s))) (define (registers-modified s) (if (symbol? s) '() (cadr s))) (define (statements s) (if (symbol? s) (list s) (caddr s)))
and to determine whether a given sequence needs or modifies a given register we use the predicates
(define (needs-register? seq reg) (memq reg (registers-needed seq))) (define (modifies-register? seq reg) (memq reg (registers-modified seq)))
In terms of these predicates and selectors, we can implement the various instruction sequence combiners used throughout the compiler.
The basic combiner is append-instruction-sequences
. This takes as
arguments an arbitrary number of instruction sequences that are to be executed
sequentially and returns an instruction sequence whose statements are the
statements of all the sequences appended together. The subtle point is to
determine the registers that are needed and modified by the resulting sequence.
It modifies those registers that are modified by any of the sequences; it needs
those registers that must be initialized before the first sequence can be run
(the registers needed by the first sequence), together with those registers
needed by any of the other sequences that are not initialized (modified) by
sequences preceding it.
The sequences are appended two at a time by append-2-sequences
. This
takes two instruction sequences seq1
and seq2
and returns the
instruction sequence whose statements are the statements of seq1
followed by the statements of seq2
, whose modified registers are those
registers that are modified by either seq1
or seq2
, and whose
needed registers are the registers needed by seq1
together with those
registers needed by seq2
that are not modified by seq1
. (In
terms of set operations, the new set of needed registers is the union of the
set of registers needed by seq1
with the set difference of the registers
needed by seq2
and the registers modified by seq1
.) Thus,
append-instruction-sequences
is implemented as follows:
(define (append-instruction-sequences . seqs) (define (append-2-sequences seq1 seq2) (make-instruction-sequence (list-union (registers-needed seq1) (list-difference (registers-needed seq2) (registers-modified seq1))) (list-union (registers-modified seq1) (registers-modified seq2)) (append (statements seq1) (statements seq2)))) (define (append-seq-list seqs) (if (null? seqs) (empty-instruction-sequence) (append-2-sequences (car seqs) (append-seq-list (cdr seqs))))) (append-seq-list seqs))
This procedure uses some simple operations for manipulating sets represented as lists, similar to the (unordered) set representation described in 2.3.3:
(define (list-union s1 s2) (cond ((null? s1) s2) ((memq (car s1) s2) (list-union (cdr s1) s2)) (else (cons (car s1) (list-union (cdr s1) s2))))) (define (list-difference s1 s2) (cond ((null? s1) '()) ((memq (car s1) s2) (list-difference (cdr s1) s2)) (else (cons (car s1) (list-difference (cdr s1) s2)))))
Preserving
, the second major instruction sequence combiner, takes a list
of registers regs
and two instruction sequences seq1
and
seq2
that are to be executed sequentially. It returns an instruction
sequence whose statements are the statements of seq1
followed by the
statements of seq2
, with appropriate save
and restore
instructions around seq1
to protect the registers in regs
that
are modified by seq1
but needed by seq2
. To accomplish this,
preserving
first creates a sequence that has the required save
s
followed by the statements of seq1
followed by the required
restore
s. This sequence needs the registers being saved and restored in
addition to the registers needed by seq1
, and modifies the registers
modified by seq1
except for the ones being saved and restored. This
augmented sequence and seq2
are then appended in the usual way. The
following procedure implements this strategy recursively, walking down the list
of registers to be preserved:327
(define (preserving regs seq1 seq2) (if (null? regs) (append-instruction-sequences seq1 seq2) (let ((first-reg (car regs))) (if (and (needs-register? seq2 first-reg) (modifies-register? seq1 first-reg)) (preserving (cdr regs) (make-instruction-sequence (list-union (list first-reg) (registers-needed seq1)) (list-difference (registers-modified seq1) (list first-reg)) (append `((save ,first-reg)) (statements seq1) `((restore ,first-reg)))) seq2) (preserving (cdr regs) seq1 seq2)))))
Another sequence combiner, tack-on-instruction-sequence
, is used by
compile-lambda
to append a procedure body to another sequence. Because
the procedure body is not “in line” to be executed as part of the combined
sequence, its register use has no impact on the register use of the sequence in
which it is embedded. We thus ignore the procedure body’s sets of needed and
modified registers when we tack it onto the other sequence.
(define (tack-on-instruction-sequence seq body-seq) (make-instruction-sequence (registers-needed seq) (registers-modified seq) (append (statements seq) (statements body-seq))))
Compile-if
and compile-procedure-call
use a special combiner
called parallel-instruction-sequences
to append the two alternative
branches that follow a test. The two branches will never be executed
sequentially; for any particular evaluation of the test, one branch or the
other will be entered. Because of this, the registers needed by the second
branch are still needed by the combined sequence, even if these are modified by
the first branch.
(define (parallel-instruction-sequences seq1 seq2) (make-instruction-sequence (list-union (registers-needed seq1) (registers-needed seq2)) (list-union (registers-modified seq1) (registers-modified seq2)) (append (statements seq1) (statements seq2))))
Now that we have seen all the elements of the compiler, let us examine an
example of compiled code to see how things fit together. We will compile the
definition of a recursive factorial
procedure by calling compile
:
(compile '(define (factorial n) (if (= n 1) 1 (* (factorial (- n 1)) n))) 'val 'next)
We have specified that the value of the define
expression should be
placed in the val
register. We don’t care what the compiled code does
after executing the define
, so our choice of next
as the linkage
descriptor is arbitrary.
Compile
determines that the expression is a definition, so it calls
compile-definition
to compile code to compute the value to be assigned
(targeted to val
), followed by code to install the definition, followed
by code to put the value of the define
(which is the symbol ok
)
into the target register, followed finally by the linkage code. Env
is
preserved around the computation of the value, because it is needed in order to
install the definition. Because the linkage is next
, there is no
linkage code in this case. The skeleton of the compiled code is thus
⟨saveenv
if modified by code to compute value⟩ ⟨compilation of definition value, targetval
, linkagenext
⟩ ⟨restoreenv
if saved above⟩ (perform (op define-variable!) (const factorial) (reg val) (reg env)) (assign val (const ok))
The expression that is to be compiled to produce the value for the variable
factorial
is a lambda
expression whose value is the procedure
that computes factorials. Compile
handles this by calling
compile-lambda
, which compiles the procedure body, labels it as a new
entry point, and generates the instruction that will combine the procedure body
at the new entry point with the run-time environment and assign the result to
val
. The sequence then skips around the compiled procedure code, which
is inserted at this point. The procedure code itself begins by extending the
procedure’s definition environment by a frame that binds the formal parameter
n
to the procedure argument. Then comes the actual procedure body.
Since this code for the value of the variable doesn’t modify the env
register, the optional save
and restore
shown above aren’t
generated. (The procedure code at entry2
isn’t executed at this point,
so its use of env
is irrelevant.) Therefore, the skeleton for the
compiled code becomes
(assign val (op make-compiled-procedure) (label entry2) (reg env)) (goto (label after-lambda1)) entry2 (assign env (op compiled-procedure-env) (reg proc)) (assign env (op extend-environment) (const (n)) (reg argl) (reg env)) ⟨compilation of procedure body⟩ after-lambda1 (perform (op define-variable!) (const factorial) (reg val) (reg env)) (assign val (const ok))
A procedure body is always compiled (by compile-lambda-body
) as a
sequence with target val
and linkage return
. The sequence in
this case consists of a single if
expression:
(if (= n 1) 1 (* (factorial (- n 1)) n))
Compile-if
generates code that first computes the predicate (targeted to
val
), then checks the result and branches around the true branch if the
predicate is false. Env
and continue
are preserved around the
predicate code, since they may be needed for the rest of the if
expression. Since the if
expression is the final expression (and only
expression) in the sequence making up the procedure body, its target is
val
and its linkage is return
, so the true and false branches are
both compiled with target val
and linkage return
. (That is, the
value of the conditional, which is the value computed by either of its
branches, is the value of the procedure.)
⟨savecontinue
,env
if modified by predicate and needed by branches⟩ ⟨compilation of predicate, targetval
, linkagenext
⟩ ⟨restorecontinue
,env
if saved above⟩ (test (op false?) (reg val)) (branch (label false-branch4)) true-branch5 ⟨compilation of true branch, targetval
, linkagereturn
⟩ false-branch4 ⟨compilation of false branch, targetval
, linkagereturn
⟩ after-if3
The predicate (= n 1)
is a procedure call. This looks up the operator
(the symbol =
) and places this value in proc
. It then assembles
the arguments 1
and the value of n
into argl
. Then it
tests whether proc
contains a primitive or a compound procedure, and
dispatches to a primitive branch or a compound branch accordingly. Both
branches resume at the after-call
label. The requirements to preserve
registers around the evaluation of the operator and operands don’t result in
any saving of registers, because in this case those evaluations don’t modify
the registers in question.
(assign proc (op lookup-variable-value) (const =) (reg env)) (assign val (const 1)) (assign argl (op list) (reg val)) (assign val (op lookup-variable-value) (const n) (reg env)) (assign argl (op cons) (reg val) (reg argl)) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch17)) compiled-branch16 (assign continue (label after-call15)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch17 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call15
The true branch, which is the constant 1, compiles (with target val
and
linkage return
) to
(assign val (const 1)) (goto (reg continue))
The code for the false branch is another procedure call, where the procedure
is the value of the symbol *
, and the arguments are n
and the
result of another procedure call (a call to factorial
). Each of these
calls sets up proc
and argl
and its own primitive and compound
branches. Figure 5.17 shows the complete compilation of the definition
of the factorial
procedure. Notice that the possible save
and
restore
of continue
and env
around the predicate, shown
above, are in fact generated, because these registers are modified by the
procedure call in the predicate and needed for the procedure call and the
return
linkage in the branches.
Figure 5.17: Compilation of the definition of the
factorial
procedure.;; construct the procedure and skip over code ;; for the procedure body (assign val (op make-compiled-procedure) (label entry2) (reg env)) (goto (label after-lambda1)) entry2 ; calls tofactorial
will enter here (assign env (op compiled-procedure-env) (reg proc)) (assign env (op extend-environment) (const (n)) (reg argl) (reg env)) ;; begin actual procedure body (save continue) (save env) ;; compute(= n 1)
(assign proc (op lookup-variable-value) (const =) (reg env)) (assign val (const 1)) (assign argl (op list) (reg val)) (assign val (op lookup-variable-value) (const n) (reg env)) (assign argl (op cons) (reg val) (reg argl)) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch17)) compiled-branch16 (assign continue (label after-call15)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch17 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call15 ;val
now contains result of(= n 1)
(restore env) (restore continue) (test (op false?) (reg val)) (branch (label false-branch4)) true-branch5 ; return 1 (assign val (const 1)) (goto (reg continue)) false-branch4 ;; compute and return(* (factorial (- n 1)) n)
(assign proc (op lookup-variable-value) (const *) (reg env)) (save continue) (save proc) ; save*
procedure (assign val (op lookup-variable-value) (const n) (reg env)) (assign argl (op list) (reg val)) (save argl) ; save partial argument list for*
;; compute(factorial (- n 1))
, ;; which is the other argument for*
(assign proc (op lookup-variable-value) (const factorial) (reg env)) (save proc) ; savefactorial
procedure ;; compute(- n 1)
, which is the argument forfactorial
(assign proc (op lookup-variable-value) (const -) (reg env)) (assign val (const 1)) (assign argl (op list) (reg val)) (assign val (op lookup-variable-value) (const n) (reg env)) (assign argl (op cons) (reg val) (reg argl)) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch8)) compiled-branch7 (assign continue (label after-call6)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch8 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call6 ;val
now contains result of(- n 1)
(assign argl (op list) (reg val)) (restore proc) ; restorefactorial
;; applyfactorial
(test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch11)) compiled-branch10 (assign continue (label after-call9)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch11 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call9 ;val
now contains result ; of(factorial (- n 1))
(restore argl) ; restore partial argument list for*
(assign argl (op cons) (reg val) (reg argl)) (restore proc) ; restore*
(restore continue) ;; apply*
and return its value (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch14)) compiled-branch13 ;; note that a compound procedure here ;; is called tail-recursively (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch14 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) (goto (reg continue)) after-call12 after-if3 after-lambda1 ;; assign the procedure to the variablefactorial
(perform (op define-variable!) (const factorial) (reg val) (reg env)) (assign val (const ok))
Exercise 5.33: Consider the following definition of a factorial procedure, which is slightly different from the one given above:
(define (factorial-alt n) (if (= n 1) 1 (* n (factorial-alt (- n 1)))))Compile this procedure and compare the resulting code with that produced for
factorial
. Explain any differences you find. Does either program execute more efficiently than the other?
Exercise 5.34: Compile the iterative factorial procedure
(define (factorial n) (define (iter product counter) (if (> counter n) product (iter (* counter product) (+ counter 1)))) (iter 1 1))Annotate the resulting code, showing the essential difference between the code for iterative and recursive versions of
factorial
that makes one process build up stack space and the other run in constant stack space.
Exercise 5.35: What expression was compiled to produce the code shown in Figure 5.18?
Figure 5.18: An example of compiler output. See Exercise 5.35.
(assign val (op make-compiled-procedure) (label entry16) (reg env)) (goto (label after-lambda15)) entry16 (assign env (op compiled-procedure-env) (reg proc)) (assign env (op extend-environment) (const (x)) (reg argl) (reg env)) (assign proc (op lookup-variable-value) (const +) (reg env)) (save continue) (save proc) (save env) (assign proc (op lookup-variable-value) (const g) (reg env)) (save proc) (assign proc (op lookup-variable-value) (const +) (reg env)) (assign val (const 2)) (assign argl (op list) (reg val)) (assign val (op lookup-variable-value) (const x) (reg env)) (assign argl (op cons) (reg val) (reg argl)) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch19)) compiled-branch18 (assign continue (label after-call17)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch19 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call17 (assign argl (op list) (reg val)) (restore proc) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch22)) compiled-branch21 (assign continue (label after-call20)) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch22 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) after-call20 (assign argl (op list) (reg val)) (restore env) (assign val (op lookup-variable-value) (const x) (reg env)) (assign argl (op cons) (reg val) (reg argl)) (restore proc) (restore continue) (test (op primitive-procedure?) (reg proc)) (branch (label primitive-branch25)) compiled-branch24 (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val)) primitive-branch25 (assign val (op apply-primitive-procedure) (reg proc) (reg argl)) (goto (reg continue)) after-call23 after-lambda15 (perform (op define-variable!) (const f) (reg val) (reg env)) (assign val (const ok))
Exercise 5.36: What order of evaluation does our compiler produce for operands of a combination? Is it left-to-right, right-to-left, or some other order? Where in the compiler is this order determined? Modify the compiler so that it produces some other order of evaluation. (See the discussion of order of evaluation for the explicit-control evaluator in 5.4.1.) How does changing the order of operand evaluation affect the efficiency of the code that constructs the argument list?
Exercise 5.37: One way to understand the compiler’s
preserving
mechanism for optimizing stack usage is to see what extra operations would be generated if we did not use this idea. Modifypreserving
so that it always generates thesave
andrestore
operations. Compile some simple expressions and identify the unnecessary stack operations that are generated. Compare the code to that generated with thepreserving
mechanism intact.
Exercise 5.38: Our compiler is clever about avoiding unnecessary stack operations, but it is not clever at all when it comes to compiling calls to the primitive procedures of the language in terms of the primitive operations supplied by the machine. For example, consider how much code is compiled to compute
(+ a 1)
: The code sets up an argument list inargl
, puts the primitive addition procedure (which it finds by looking up the symbol+
in the environment) intoproc
, and tests whether the procedure is primitive or compound. The compiler always generates code to perform the test, as well as code for primitive and compound branches (only one of which will be executed). We have not shown the part of the controller that implements primitives, but we presume that these instructions make use of primitive arithmetic operations in the machine’s data paths. Consider how much less code would be generated if the compiler could open-code primitives—that is, if it could generate code to directly use these primitive machine operations. The expression(+ a 1)
might be compiled into something as simple as328(assign val (op lookup-variable-value) (const a) (reg env)) (assign val (op +) (reg val) (const 1))In this exercise we will extend our compiler to support open coding of selected primitives. Special-purpose code will be generated for calls to these primitive procedures instead of the general procedure-application code. In order to support this, we will augment our machine with special argument registers
arg1
andarg2
. The primitive arithmetic operations of the machine will take their inputs fromarg1
andarg2
. The results may be put intoval
,arg1
, orarg2
.The compiler must be able to recognize the application of an open-coded primitive in the source program. We will augment the dispatch in the
compile
procedure to recognize the names of these primitives in addition to the reserved words (the special forms) it currently recognizes.329 For each special form our compiler has a code generator. In this exercise we will construct a family of code generators for the open-coded primitives.
- The open-coded primitives, unlike the special forms, all need their operands evaluated. Write a code generator
spread-arguments
for use by all the open-coding code generators.Spread-arguments
should take an operand list and compile the given operands targeted to successive argument registers. Note that an operand may contain a call to an open-coded primitive, so argument registers will have to be preserved during operand evaluation.- For each of the primitive procedures
=
,*
,-
, and+
, write a code generator that takes a combination with that operator, together with a target and a linkage descriptor, and produces code to spread the arguments into the registers and then perform the operation targeted to the given target with the given linkage. You need only handle expressions with two operands. Makecompile
dispatch to these code generators.- Try your new compiler on the
factorial
example. Compare the resulting code with the result produced without open coding.- Extend your code generators for
+
and*
so that they can handle expressions with arbitrary numbers of operands. An expression with more than two operands will have to be compiled into a sequence of operations, each with only two inputs.
One of the most common optimizations performed by compilers is the optimization
of variable lookup. Our compiler, as we have implemented it so far, generates
code that uses the lookup-variable-value
operation of the evaluator
machine. This searches for a variable by comparing it with each variable that
is currently bound, working frame by frame outward through the run-time
environment. This search can be expensive if the frames are deeply nested or
if there are many variables. For example, consider the problem of looking up
the value of x
while evaluating the expression (* x y z)
in an
application of the procedure that is returned by
(let ((x 3) (y 4)) (lambda (a b c d e) (let ((y (* a b x)) (z (+ c d x))) (* x y z))))
Since a let
expression is just syntactic sugar for a lambda
combination, this expression is equivalent to
((lambda (x y) (lambda (a b c d e) ((lambda (y z) (* x y z)) (* a b x) (+ c d x)))) 3 4)
Each time lookup-variable-value
searches for x
, it must determine
that the symbol x
is not eq?
to y
or z
(in the
first frame), nor to a
, b
, c
, d
, or e
(in
the second frame). We will assume, for the moment, that our programs do not
use define
—that variables are bound only with lambda
. Because
our language is lexically scoped, the run-time environment for any expression
will have a structure that parallels the lexical structure of the program in
which the expression appears.330 Thus, the
compiler can know, when it analyzes the above expression, that each time the
procedure is applied the variable x
in (* x y z)
will be found
two frames out from the current frame and will be the first variable in that
frame.
We can exploit this fact by inventing a new kind of variable-lookup operation,
lexical-address-lookup
, that takes as arguments an environment and a
lexical address that consists of two numbers: a
frame number,
which specifies how many frames to pass over, and a
displacement number, which specifies how many variables to pass over
in that frame. Lexical-address-lookup
will produce the value of the
variable stored at that lexical address relative to the current environment.
If we add the lexical-address-lookup
operation to our machine, we can
make the compiler generate code that references variables using this operation,
rather than lookup-variable-value
. Similarly, our compiled code can use
a new lexical-address-set!
operation instead of
set-variable-value!
.
In order to generate such code, the compiler must be able to determine the
lexical address of a variable it is about to compile a reference to. The
lexical address of a variable in a program depends on where one is in the code.
For example, in the following program, the address of x
in expression
⟨
e1⟩
is (2, 0)—two frames back and the first variable in the frame. At
that point y
is at address (0, 0) and c
is at address (1, 2). In
expression ⟨
e2⟩
, x
is at (1, 0), y
is at (1, 1), and c
is at (0, 2).
((lambda (x y) (lambda (a b c d e) ((lambda (y z) ⟨e1⟩) ⟨e2⟩ (+ c d x)))) 3 4)
One way for the compiler to produce code that uses lexical addressing is to
maintain a data structure called a
compile-time environment. This
keeps track of which variables will be at which positions in which frames in
the run-time environment when a particular variable-access operation is
executed. The compile-time environment is a list of frames, each containing a
list of variables. (There will of course be no values bound to the variables,
since values are not computed at compile time.) The compile-time environment
becomes an additional argument to compile
and is passed along to each
code generator. The top-level call to compile
uses an empty
compile-time environment. When a lambda
body is compiled,
compile-lambda-body
extends the compile-time environment by a frame
containing the procedure’s parameters, so that the sequence making up the body
is compiled with that extended environment. At each point in the compilation,
compile-variable
and compile-assignment
use the compile-time
environment in order to generate the appropriate lexical addresses.
Exercise 5.39 through Exercise 5.43 describe how to complete this sketch of the lexical-addressing strategy in order to incorporate lexical lookup into the compiler. Exercise 5.44 describes another use for the compile-time environment.
Exercise 5.39: Write a procedure
lexical-address-lookup
that implements the new lookup operation. It should take two arguments—a lexical address and a run-time environment—and return the value of the variable stored at the specified lexical address.Lexical-address-lookup
should signal an error if the value of the variable is the symbol*unassigned*
.331 Also write a procedurelexical-address-set!
that implements the operation that changes the value of the variable at a specified lexical address.
Exercise 5.40: Modify the compiler to maintain the compile-time environment as described above. That is, add a compile-time-environment argument to
compile
and the various code generators, and extend it incompile-lambda-body
.
Exercise 5.41: Write a procedure
find-variable
that takes as arguments a variable and a compile-time environment and returns the lexical address of the variable with respect to that environment. For example, in the program fragment that is shown above, the compile-time environment during the compilation of expression⟨
e1⟩
is((y z) (a b c d e) (x y))
.Find-variable
should produce(find-variable 'c '((y z) (a b c d e) (x y))) (1 2) (find-variable 'x '((y z) (a b c d e) (x y))) (2 0) (find-variable 'w '((y z) (a b c d e) (x y))) not-found
Exercise 5.42: Using
find-variable
from Exercise 5.41, rewritecompile-variable
andcompile-assignment
to output lexical-address instructions. In cases wherefind-variable
returnsnot-found
(that is, where the variable is not in the compile-time environment), you should have the code generators use the evaluator operations, as before, to search for the binding. (The only place a variable that is not found at compile time can be is in the global environment, which is part of the run-time environment but is not part of the compile-time environment.332 Thus, if you wish, you may have the evaluator operations look directly in the global environment, which can be obtained with the operation(op get-global-environment)
, instead of having them search the whole run-time environment found inenv
.) Test the modified compiler on a few simple cases, such as the nestedlambda
combination at the beginning of this section.
Exercise 5.43: We argued in 4.1.6 that internal definitions for block structure should not be considered “real”
define
s. Rather, a procedure body should be interpreted as if the internal variables being defined were installed as ordinarylambda
variables initialized to their correct values usingset!
. 4.1.6 and Exercise 4.16 showed how to modify the metacircular interpreter to accomplish this by scanning out internal definitions. Modify the compiler to perform the same transformation before it compiles a procedure body.
Exercise 5.44: In this section we have focused on the use of the compile-time environment to produce lexical addresses. But there are other uses for compile-time environments. For instance, in Exercise 5.38 we increased the efficiency of compiled code by open-coding primitive procedures. Our implementation treated the names of open-coded procedures as reserved words. If a program were to rebind such a name, the mechanism described in Exercise 5.38 would still open-code it as a primitive, ignoring the new binding. For example, consider the procedure
(lambda (+ * a b x y) (+ (* a x) (* b y)))which computes a linear combination of
x
andy
. We might call it with arguments+matrix
,*matrix
, and four matrices, but the open-coding compiler would still open-code the+
and the*
in(+ (* a x) (* b y))
as primitive+
and*
. Modify the open-coding compiler to consult the compile-time environment in order to compile the correct code for expressions involving the names of primitive procedures. (The code will work correctly as long as the program does notdefine
orset!
these names.)
We have not yet explained how to load compiled code into the evaluator machine
or how to run it. We will assume that the explicit-control-evaluator machine
has been defined as in 5.4.4, with the additional operations
specified in Footnote 323. We will implement a procedure
compile-and-go
that compiles a Scheme expression, loads the resulting
object code into the evaluator machine, and causes the machine to run the code
in the evaluator global environment, print the result, and enter the
evaluator’s driver loop. We will also modify the evaluator so that interpreted
expressions can call compiled procedures as well as interpreted ones. We can
then put a compiled procedure into the machine and use the evaluator to call
it:
(compile-and-go '(define (factorial n) (if (= n 1) 1 (* (factorial (- n 1)) n)))) ;;; EC-Eval value: ok ;;; EC-Eval input: (factorial 5) ;;; EC-Eval value: 120
To allow the evaluator to handle compiled procedures (for example, to evaluate
the call to factorial
above), we need to change the code at
apply-dispatch
(5.4.1) so that it recognizes compiled
procedures (as distinct from compound or primitive procedures) and transfers
control directly to the entry point of the compiled code:333
apply-dispatch (test (op primitive-procedure?) (reg proc)) (branch (label primitive-apply)) (test (op compound-procedure?) (reg proc)) (branch (label compound-apply)) (test (op compiled-procedure?) (reg proc)) (branch (label compiled-apply)) (goto (label unknown-procedure-type)) compiled-apply (restore continue) (assign val (op compiled-procedure-entry) (reg proc)) (goto (reg val))
Note the restore of continue
at compiled-apply
. Recall that the
evaluator was arranged so that at apply-dispatch
, the continuation would
be at the top of the stack. The compiled code entry point, on the other hand,
expects the continuation to be in continue
, so continue
must be
restored before the compiled code is executed.
To enable us to run some compiled code when we start the evaluator machine, we
add a branch
instruction at the beginning of the evaluator machine,
which causes the machine to go to a new entry point if the flag
register
is set.334
;; branches if flag
is set:
(branch (label external-entry))
read-eval-print-loop
(perform (op initialize-stack))
…
External-entry
assumes that the machine is started with val
containing the location of an instruction sequence that puts a result into
val
and ends with (goto (reg continue))
. Starting at this entry
point jumps to the location designated by val
, but first assigns
continue
so that execution will return to print-result
, which
prints the value in val
and then goes to the beginning of the
evaluator’s read-eval-print loop.335
external-entry (perform (op initialize-stack)) (assign env (op get-global-environment)) (assign continue (label print-result)) (goto (reg val))
Now we can use the following procedure to compile a procedure definition,
execute the compiled code, and run the read-eval-print loop so we can try the
procedure. Because we want the compiled code to return to the location in
continue
with its result in val
, we compile the expression with a
target of val
and a linkage of return
. In order to transform the
object code produced by the compiler into executable instructions for the
evaluator register machine, we use the procedure assemble
from the
register-machine simulator (5.2.2). We then initialize the
val
register to point to the list of instructions, set the flag
so that the evaluator will go to external-entry
, and start the
evaluator.
(define (compile-and-go expression) (let ((instructions (assemble (statements (compile expression 'val 'return)) eceval))) (set! the-global-environment (setup-environment)) (set-register-contents! eceval 'val instructions) (set-register-contents! eceval 'flag true) (start eceval)))
If we have set up stack monitoring, as at the end of 5.4.4, we can examine the stack usage of compiled code:
(compile-and-go '(define (factorial n) (if (= n 1) 1 (* (factorial (- n 1)) n)))) (total-pushes = 0, maximum-depth = 0) ;;; EC-Eval value: ok ;;; EC-Eval input: (factorial 5) (total-pushes = 31, maximum-depth = 14) ;;; EC-Eval value: 120
Compare this example with the evaluation of (factorial 5)
using the
interpreted version of the same procedure, shown at the end of
5.4.4. The interpreted version required 144 pushes and a maximum stack
depth of 28. This illustrates the optimization that results from our
compilation strategy.
With the programs in this section, we can now experiment with the alternative execution strategies of interpretation and compilation.336 An interpreter raises the machine to the level of the user program; a compiler lowers the user program to the level of the machine language. We can regard the Scheme language (or any programming language) as a coherent family of abstractions erected on the machine language. Interpreters are good for interactive program development and debugging because the steps of program execution are organized in terms of these abstractions, and are therefore more intelligible to the programmer. Compiled code can execute faster, because the steps of program execution are organized in terms of the machine language, and the compiler is free to make optimizations that cut across the higher-level abstractions.337
The alternatives of interpretation and compilation also lead to different strategies for porting languages to new computers. Suppose that we wish to implement Lisp for a new machine. One strategy is to begin with the explicit-control evaluator of 5.4 and translate its instructions to instructions for the new machine. A different strategy is to begin with the compiler and change the code generators so that they generate code for the new machine. The second strategy allows us to run any Lisp program on the new machine by first compiling it with the compiler running on our original Lisp system, and linking it with a compiled version of the run-time library.338 Better yet, we can compile the compiler itself, and run this on the new machine to compile other Lisp programs.339 Or we can compile one of the interpreters of 4.1 to produce an interpreter that runs on the new machine.
Exercise 5.45: By comparing the stack operations used by compiled code to the stack operations used by the evaluator for the same computation, we can determine the extent to which the compiler optimizes use of the stack, both in speed (reducing the total number of stack operations) and in space (reducing the maximum stack depth). Comparing this optimized stack use to the performance of a special-purpose machine for the same computation gives some indication of the quality of the compiler.
- Exercise 5.27 asked you to determine, as a function of , the number of pushes and the maximum stack depth needed by the evaluator to compute using the recursive factorial procedure given above. Exercise 5.14 asked you to do the same measurements for the special-purpose factorial machine shown in Figure 5.11. Now perform the same analysis using the compiled
factorial
procedure.Take the ratio of the number of pushes in the compiled version to the number of pushes in the interpreted version, and do the same for the maximum stack depth. Since the number of operations and the stack depth used to compute are linear in , these ratios should approach constants as becomes large. What are these constants? Similarly, find the ratios of the stack usage in the special-purpose machine to the usage in the interpreted version.
Compare the ratios for special-purpose versus interpreted code to the ratios for compiled versus interpreted code. You should find that the special-purpose machine does much better than the compiled code, since the hand-tailored controller code should be much better than what is produced by our rudimentary general-purpose compiler.
- Can you suggest improvements to the compiler that would help it generate code that would come closer in performance to the hand-tailored version?
Exercise 5.46: Carry out an analysis like the one in Exercise 5.45 to determine the effectiveness of compiling the tree-recursive Fibonacci procedure
(define (fib n) (if (< n 2) n (+ (fib (- n 1)) (fib (- n 2)))))compared to the effectiveness of using the special-purpose Fibonacci machine of Figure 5.12. (For measurement of the interpreted performance, see Exercise 5.29.) For Fibonacci, the time resource used is not linear in hence the ratios of stack operations will not approach a limiting value that is independent of .
Exercise 5.47: This section described how to modify the explicit-control evaluator so that interpreted code can call compiled procedures. Show how to modify the compiler so that compiled procedures can call not only primitive procedures and compiled procedures, but interpreted procedures as well. This requires modifying
compile-procedure-call
to handle the case of compound (interpreted) procedures. Be sure to handle all the sametarget
andlinkage
combinations as incompile-proc-appl
. To do the actual procedure application, the code needs to jump to the evaluator’scompound-apply
entry point. This label cannot be directly referenced in object code (since the assembler requires that all labels referenced by the code it is assembling be defined there), so we will add a register calledcompapp
to the evaluator machine to hold this entry point, and add an instruction to initialize it:(assign compapp (label compound-apply)) ;; branches ifflag
is set: (branch (label external-entry)) read-eval-print-loop …To test your code, start by defining a procedure
f
that calls a procedureg
. Usecompile-and-go
to compile the definition off
and start the evaluator. Now, typing at the evaluator, defineg
and try to callf
.
Exercise 5.48: The
compile-and-go
interface implemented in this section is awkward, since the compiler can be called only once (when the evaluator machine is started). Augment the compiler-interpreter interface by providing acompile-and-run
primitive that can be called from within the explicit-control evaluator as follows:;;; EC-Eval input: (compile-and-run '(define (factorial n) (if (= n 1) 1 (* (factorial (- n 1)) n)))) ;;; EC-Eval value: ok ;;; EC-Eval input: (factorial 5) ;;; EC-Eval value: 120
Exercise 5.49: As an alternative to using the explicit-control evaluator’s read-eval-print loop, design a register machine that performs a read-compile-execute-print loop. That is, the machine should run a loop that reads an expression, compiles it, assembles and executes the resulting code, and prints the result. This is easy to run in our simulated setup, since we can arrange to call the procedures
compile
andassemble
as “register-machine operations.”
Exercise 5.50: Use the compiler to compile the metacircular evaluator of 4.1 and run this program using the register-machine simulator. (To compile more than one definition at a time, you can package the definitions in a
begin
.) The resulting interpreter will run very slowly because of the multiple levels of interpretation, but getting all the details to work is an instructive exercise.
Exercise 5.51: Develop a rudimentary implementation of Scheme in C (or some other low-level language of your choice) by translating the explicit-control evaluator of 5.4 into C. In order to run this code you will need to also provide appropriate storage-allocation routines and other run-time support.
Exercise 5.52: As a counterpoint to Exercise 5.51, modify the compiler so that it compiles Scheme procedures into sequences of C instructions. Compile the metacircular evaluator of 4.1 to produce a Scheme interpreter written in C.
318 This is a theoretical statement. We are not claiming that the evaluator’s data paths are a particularly convenient or efficient set of data paths for a general-purpose computer. For example, they are not very good for implementing high-performance floating-point calculations or calculations that intensively manipulate bit vectors.
319
Actually, the machine that runs compiled code can be simpler
than the interpreter machine, because we won’t use the exp
and
unev
registers. The interpreter used these to hold pieces of
unevaluated expressions. With the compiler, however, these expressions get
built into the compiled code that the register machine will run. For the same
reason, we don’t need the machine operations that deal with expression syntax.
But compiled code will use a few additional machine operations (to represent
compiled procedure objects) that didn’t appear in the explicit-control
evaluator machine.
320 Notice, however, that our compiler is a Scheme program, and the syntax procedures that it uses to manipulate expressions are the actual Scheme procedures used with the metacircular evaluator. For the explicit-control evaluator, in contrast, we assumed that equivalent syntax operations were available as operations for the register machine. (Of course, when we simulated the register machine in Scheme, we used the actual Scheme procedures in our register machine simulation.)
321 This procedure uses a feature of Lisp called backquote (or quasiquote) that is handy for constructing lists. Preceding a list with a backquote symbol is much like quoting it, except that anything in the list that is flagged with a comma is evaluated.
For example, if the value of linkage
is the symbol branch25
,
then the expression
`((goto (label ,linkage)))
evaluates to the list
((goto (label branch25)))
Similarly, if the value of x
is the list (a b c)
, then
`(1 2 ,(car x))
evaluates to the list
(1 2 a)
322
We can’t just use the labels true-branch
,
false-branch
, and after-if
as shown above, because there might be
more than one if
in the program. The compiler uses the procedure
make-label
to generate labels. Make-label
takes a symbol as
argument and returns a new symbol that begins with the given symbol. For
example, successive calls to (make-label 'a)
would return a1
,
a2
, and so on. Make-label
can be implemented similarly to the
generation of unique variable names in the query language, as follows:
(define label-counter 0) (define (new-label-number) (set! label-counter (+ 1 label-counter)) label-counter) (define (make-label name) (string->symbol (string-append (symbol->string name) (number->string (new-label-number)))))
323 We need machine operations to implement a data structure for representing compiled procedures, analogous to the structure for compound procedures described in 4.1.3:
(define (make-compiled-procedure entry env) (list 'compiled-procedure entry env)) (define (compiled-procedure? proc) (tagged-list? proc 'compiled-procedure)) (define (compiled-procedure-entry c-proc) (cadr c-proc)) (define (compiled-procedure-env c-proc) (caddr c-proc))
324
Actually, we signal an error when the target is not
val
and the linkage is return
, since the only place we request
return
linkages is in compiling procedures, and our convention is that
procedures return their values in val
.
325 Making a compiler generate tail-recursive code might seem like a straightforward idea. But most compilers for common languages, including C and Pascal, do not do this, and therefore these languages cannot represent iterative processes in terms of procedure call alone. The difficulty with tail recursion in these languages is that their implementations use the stack to store procedure arguments and local variables as well as return addresses. The Scheme implementations described in this book store arguments and variables in memory to be garbage-collected. The reason for using the stack for variables and arguments is that it avoids the need for garbage collection in languages that would not otherwise require it, and is generally believed to be more efficient. Sophisticated Lisp compilers can, in fact, use the stack for arguments without destroying tail recursion. (See Hanson 1990 for a description.) There is also some debate about whether stack allocation is actually more efficient than garbage collection in the first place, but the details seem to hinge on fine points of computer architecture. (See Appel 1987 and Miller and Rozas 1994 for opposing views on this issue.)
326
The
variable all-regs
is bound to the list of names of all the registers:
(define all-regs '(env proc val argl continue))
327
Note that preserving
calls
append
with three arguments. Though the definition of append
shown in this book accepts only two arguments, Scheme standardly provides an
append
procedure that takes an arbitrary number of arguments.
328
We have used the same symbol
+
here to denote both the source-language procedure and the machine
operation. In general there will not be a one-to-one correspondence between
primitives of the source language and primitives of the machine.
329 Making the primitives into reserved words is in general a bad idea, since a user cannot then rebind these names to different procedures. Moreover, if we add reserved words to a compiler that is in use, existing programs that define procedures with these names will stop working. See Exercise 5.44 for ideas on how to avoid this problem.
330 This is not true if we allow internal definitions, unless we scan them out. See Exercise 5.43.
331 This is the modification to variable lookup required if we implement the scanning method to eliminate internal definitions (Exercise 5.43). We will need to eliminate these definitions in order for lexical addressing to work.
332 Lexical addresses cannot be used to access variables in the global environment, because these names can be defined and redefined interactively at any time. With internal definitions scanned out, as in Exercise 5.43, the only definitions the compiler sees are those at top level, which act on the global environment. Compilation of a definition does not cause the defined name to be entered in the compile-time environment.
333 Of course, compiled procedures as well as interpreted procedures are compound (nonprimitive). For compatibility with the terminology used in the explicit-control evaluator, in this section we will use “compound” to mean interpreted (as opposed to compiled).
334
Now that the evaluator machine starts with a branch
, we
must always initialize the flag
register before starting the evaluator
machine. To start the machine at its ordinary read-eval-print loop, we could
use
(define (start-eceval) (set! the-global-environment (setup-environment)) (set-register-contents! eceval 'flag false) (start eceval))
335
Since a compiled procedure is an
object that the system may try to print, we also modify the system print
operation user-print
(from 4.1.4) so that it will not
attempt to print the components of a compiled procedure:
(define (user-print object) (cond ((compound-procedure? object) (display (list 'compound-procedure (procedure-parameters object) (procedure-body object) '<procedure-env>))) ((compiled-procedure? object) (display '<compiled-procedure>)) (else (display object))))
336 We can do even better by extending the compiler to allow compiled code to call interpreted procedures. See Exercise 5.47.
337 Independent of the strategy of execution, we incur significant overhead if we insist that errors encountered in execution of a user program be detected and signaled, rather than being allowed to kill the system or produce wrong answers. For example, an out-of-bounds array reference can be detected by checking the validity of the reference before performing it. The overhead of checking, however, can be many times the cost of the array reference itself, and a programmer should weigh speed against safety in determining whether such a check is desirable. A good compiler should be able to produce code with such checks, should avoid redundant checks, and should allow programmers to control the extent and type of error checking in the compiled code.
Compilers for popular languages, such as C and C++, put hardly any error-checking operations into running code, so as to make things run as fast as possible. As a result, it falls to programmers to explicitly provide error checking. Unfortunately, people often neglect to do this, even in critical applications where speed is not a constraint. Their programs lead fast and dangerous lives. For example, the notorious “Worm” that paralyzed the Internet in 1988 exploited the UNIX(tm) operating system’s failure to check whether the input buffer has overflowed in the finger daemon. (See Spafford 1989.)
338 Of course, with either the interpretation or the compilation strategy we must also implement for the new machine storage allocation, input and output, and all the various operations that we took as “primitive” in our discussion of the evaluator and compiler. One strategy for minimizing work here is to write as many of these operations as possible in Lisp and then compile them for the new machine. Ultimately, everything reduces to a small kernel (such as garbage collection and the mechanism for applying actual machine primitives) that is hand-coded for the new machine.
339 This strategy leads to amusing tests of correctness of the compiler, such as checking whether the compilation of a program on the new machine, using the compiled compiler, is identical with the compilation of the program on the original Lisp system. Tracking down the source of differences is fun but often frustrating, because the results are extremely sensitive to minuscule details.