To evaluate a combination whose operator names a compound procedure, the
interpreter follows much the same process as for combinations whose
operators name primitive procedures, which we described in
section
. That is, the interpreter
evaluates the elements of the combination and applies the procedure
(which is the value of the operator of the combination) to the
arguments (which are the values of the operands of the combination).
We can assume that the mechanism for applying primitive procedures to arguments is built into the interpreter. For compound procedures, the application process is as follows:
To illustrate this process, let's evaluate the combination
(f 5)where f is the procedure defined in section
. We begin by retrieving the
body of f:
(sum-of-squares (+ a 1) (* a 2))Then we replace the formal parameter a by the argument 5:
(sum-of-squares (+ 5 1) (* 5 2))Thus the problem reduces to the evaluation of a combination with two operands and an operator sum-of-squares. Evaluating this combination involves three subproblems. We must evaluate the operator to get the procedure to be applied, and we must evaluate the operands to get the arguments. Now (+ 5 1) produces 6 and (* 5 2) produces 10, so we must apply the sum-of-squares procedure to 6 and 10. These values are substituted for the formal parameters x and y in the body of sum-of-squares, reducing the expression to
(+ (square 6) (square 10))If we use the definition of square, this reduces to
(+ (* 6 6) (* 10 10))which reduces by multiplication to
(+ 36 100)and finally to
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The process we have just described is called the substitution model for procedure application. It can be taken as a model that determines the ``meaning'' of procedure application, insofar as the procedures in this chapter are concerned. However, there are two points that should be stressed:
Over the course of this book, we will present a sequence of
increasingly elaborate models of how interpreters work, culminating
with a complete implementation of an interpreter and compiler in
chapter 5. The substitution model is only the first of these
models--a way to get started thinking formally about the evaluation
process. In general, when
modeling phenomena in science and
engineering, we begin with simplified, incomplete models. As we
examine things in greater detail, these simple models become
inadequate and must be replaced by more refined models. The
substitution model is no exception. In particular, when we address in
chapter 3 the use of procedures with ``mutable data,'' we will see that
the substitution model breaks down and must be replaced by a more
complicated model of procedure application.
Applicative order versus normal order
According to the description of evaluation given in
section
, the interpreter first
evaluates the operator and operands and then applies the resulting procedure
to the resulting arguments. This is not the only way to perform
evaluation. An alternative evaluation model would not evaluate the
operands until their values were needed. Instead it would first substitute
operand expressions for parameters until
it obtained an expression involving only primitive operators, and
would then perform the evaluation. If we used this method, the
evaluation of
(f 5)would proceed according to the sequence of expansions
(sum-of-squares (+ 5 1) (* 5 2))followed by the reductions(+ (square (+ 5 1)) (square (* 5 2)) )
(+ (* (+ 5 1) (+ 5 1)) (* (* 5 2) (* 5 2)))
(+ (* 6 6) (* 10 10))This gives the same answer as our previous evaluation model, but the process is different. In particular, the evaluations of (+ 5 1) and (* 5 2) are each performed twice here, corresponding to the reduction of the expression(+ 36 100)
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(* x x)with x replaced respectively by (+ 5 1) and (* 5 2).
This alternative ``fully expand and then reduce'' evaluation method is
known as
normal-order evaluation, in contrast to the ``evaluate
the arguments and then apply'' method that the interpreter actually
uses, which is called
applicative-order evaluation. It can be
shown that, for procedure applications that can be modeled using
substitution (including all the procedures in the first two chapters
of this book) and that yield legitimate values, normal-order and
applicative-order evaluation produce the same value. (See
exercise
for an instance of
an ``illegitimate'' value where normal-order and applicative-order
evaluation do not give the same result.)
Lisp uses applicative-order evaluation, partly because of the
additional efficiency obtained from avoiding multiple evaluations of
expressions such as those illustrated with (+ 5 1) and (* 5
2) above and, more significantly, because normal-order evaluation
becomes much more complicated to deal with when we leave the realm of
procedures that can be modeled by substitution. On the other hand,
normal-order evaluation can be an extremely valuable tool, and we will
investigate some of its implications in chapters 3 and 4.