Exercise 1.5
Given the following two procedures:
1 2 3 4 5 6 | (define (p) (p)) (define (test x y) (if (= x 0) 0 y)) |
What is the result of evaluating
1 | (test 0 (p)) |
using the following two substitution models.
Applicative-order evaluation evaluates all the expression arguments and then applies:
1 2 3 4 5 | (test 0 (p)) (test 0 (p)) . . . |
The evaluation enters into an infinite loop since the second expression argument is evaluated to itself.
Normal-order evaluation substitutes operand expressions for parameters until it obtains an expression involving only primitive operators:
1 2 3 4 5 6 7 8 9 | (test 0 (p)) (if (= 0 0) 0 (p)) (if #t 0 (p)) 0 |
Value: 0
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