SICP - Solution: Exercise 1.5

SICP - Solution: Exercise 1.5

September 27, 2018

Exercise 1.5 #

Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation. He defines the following two procedures:

(define (p) (p))

(define (test x y)
  (if (= x 0)

Then he evaluates the expression

(test 0 (p))

What behavior will Ben observe with an interpreter that uses applicative-order evaluation? What behavior will he observe with an interpreter that uses normal-order evaluation? Explain your answer.

Solution #

The key is to notice that (define (p) (p)) defines a function that evaluates to itself.

An interpreter that uses applicative-order evaluation will “evaluate the arguments and then apply”. When this kind of interpreter evaluates the expression (test 0 (p)), it will start by evaluating 0, then it will try to evaluate (p) and finally apply test to the values of the evaluation of the two parameters.

test will evaluate to the procedure defined.

0 will evaluate to the number O.

(p) will evaluate to (p), which will evaluate to (p)

To apply a compound procedure to arguments, evaluate the body of the procedure with each formal parameter replaced by the corresponding argument.

When the interpreter try to evaluate the expression (p), it will:

  1. replace each formal parameter by the corresponding argument in the body of the procedure: since there is no formal parameter in this case, the body of the procedure will just be (p).
  2. evaluated the body of the procedure, which will be (p) in our case, which in turn starts the evaluation all over again, thus making an infinite loop.

With an interpreter that uses normal-order evaluation, the interpreter will “fully expand and then reduce”. In this model, the interpreter will not evaluate the operands until their values are actually needed. In that case (test 0 (p)) will evaluate as follows:

(test 0 (p))

will be expanded to:

(if (= 0 0)

Since the predicate (= 0 0) evaluates to #t in the if, then (p) will not be evaluated and the result will be: