exercise-1-05.txt

In an interpreter that uses applicative-order evaluation, when:

(test 0 (p))

is evaluated, the interpreter will first try to evaluate the
arguments. Because the function p recurses infinitely, the interpreter
will just sit there evaluating p forever. In a normal-order
interpreter, test will first be expanded out to:

(if (= 0 0) 0 (p))

At this point the semantics of if evaluate the predicate, which is
true, then we evaluate and return the consequent 0. Because (p) is
never evaluated, we never get stuck in its infinite recursion.