Question

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

(define (improve guess x)
  (average guess(/ x guess)))

(define (average x y)
  (/ (+ x y) 2))

(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.0001))

(define (square x)
  (* x x))

(define (sqrt-g x)
  (sqrt-iter 1.0 x))

This is a program for sqrt. And the question is what happens when you attempts to use new-if to replace if with new-if.

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

This is new if

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

My opinion is the result of two program gonna be the same. because new-if and if can produce the same results.

However, new-if proved wrong, because it is a dead circle when I tried.

So, why?

Answer 1

new-if is a function. All the arguments to a function are evaluated before calling the function. But sqrt-iter is a recursive function, and you need to avoid making the recursive call when the argument is already good enough.

The built-in if is syntax, and only evaluates the then-branch or else-branch, depending on the value of the condition.

You can use a macro to write new-if.