Exercise 1.6
Given this new definition of the if special form:
1 2 3 | (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause))) |
It is used in the following program:
1 2 3 4 5 | (define (sqrt-iter guess x) (new-if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) |
What happens is that the operands of the new-if are evaluated, which means that the computation will not finish due to a recursive invocation chain of the sqrt-iter procedure, as illustrated here:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | (sqrt 4) (sqrt-iter 1.0 4) (new-if (good-enough? 1.0 4) 4 (sqrt-iter (improve 1.0 4) 4))) (new-if #f 4 (new-if (good-enough? 2.5 4) 2.5 (sqrt-iter (improve 2.5 4) 4))) (new-if #f 4 (new-if #f 2.5 (new-if (good-enough? 2.05 4) 2.05 (sqrt-iter (improve 2.05 4) 4))) . . . |
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