What's the explanation for Exercise 1.6 in SICP?

recursion conditional-statements evaluation sicp
Alex Basson picture Alex Basson · Jul 23, 2009 · Viewed 9.4k times · Source

I'm just beginning to work through SICP (on my own; this isn't for a class), and I've been struggling with Exercise 1.6 for a couple of days and I just can't seem to figure it out. This is the one where Alyssa re-defines if in terms of cond, like so:

(define (new-if predicate then-clause else-clause)
    (cond (predicate then-clause)
          (else else-clause))

She tests it successfully on some simple cases, and then uses it to re-write the square root program (which worked just fine with if):

(define (sqrt-iter guess x)
    (new-if (good-enough? guess x)
            guess
            (sqrt-iter (improve guess x)
                       x)))

The question then asks: "What happens when Alyssa attempts to use this to compute square roots? Explain." [If necessary, I'm happy to reproduce the other procedures (good-enough?, improve, etc.), just let me know.]

Now, I know what happens: it never returns a value, which means that the program recurses infinitely. I just can't explain why this happens. Whatever subtle difference exists between if and new-if is eluding me. Any and all help much appreciated.

Answer

Greg Hewgill picture Greg Hewgill · Jul 23, 2009

new-if is a function. When a function is called, what's the first thing that Scheme does with the argument list? It evaluates all the arguments.

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