SICP Exercise 1.34

Question

Suppose we define the procedure f. What happens if we (perversely) ask the interpreter to evaluate the combination (f f)?

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

(define (f g)
  (g 2))

Then we have

(f square)                          ;; 4

(f (lambda (z) (* z (+ z 1))))      ;; 6 = 2 * 3

Answer

Recall what the applicative-order evaluation says: We need to evaluate all arguments and then we apply procedure on these arguments.

The first step: we evalute the argument f in (f f).

;; replace g with f
;; (define (f g)
;;   (g 2))

(f 2)

The second step: we evaluate the argument 2 in (f 2). Note that the 2 here is regarded as a procedure rather than a primitive number.

;; replace g with 2
;; (define (f g)
;;   (g 2))

(2 2)

So we finally get the S-expression (2 2). It’s regarded as a function/procedure application. That’s how we interprete an S-expression :)

However, the 2 is not a procedure at all. This explains why Racket says:

application: not a procedure;
expected a procedure that can be applied to arguments
given: 2

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