SICP exercise 1.06

From Drewiki

Problem

Given an alternative implementation of if:

and a square-root procedure which uses it (good-enough? and improve taken from section 1.1.7):

what happens when Alyssa attempts to use the new sqrt-iter?

Solution

Consider the definition of sqrt-iter given in the text; let's call it sqrt-iter-orig here. It uses the special form if:

The if expression in sqrt-iter-orig only evaluates its else-clause expression if the predicate expression is false, so sqrt-iter-orig only performs an additional iteration (using an improved guess) if the current guess isn't good enough.

new-if isn't a special form, it's a compound procedure. The interpreter will attempt to evaluate all three of its arguments -- predicate, then-clause and else-clause -- regardless of the value of its predicate. Since evaluating the else-clause of new-if in the body of sqrt-iter requires evaluating sqrt-iter again, the interpreter will never return a value for applications of sqrt-iter. The guess will continue to improve, but there is no expression in the sqrt-iter procedure that will "short-circuit" the evaluation of subsequent guesses once the guess is good enough.