1.34 Exercise 1.34

zthomae’s SICP solutions

1

1.1	Exercise 1.1
1.2	Exercise 1.2
1.3	Exercise 1.3
1.4	Exercise 1.4
1.5	Exercise 1.5
1.6	Exercise 1.6
1.7	Exercise 1.7
1.8	Exercise 1.8
1.9	Exercise 1.9
1.10	Exercise 1.10
1.11	Exercise 1.11
1.12	Exercise 1.12
1.13	Exercise 1.13
1.14	Exercise 1.14
1.15	Exercise 1.15
1.16	Exercise 1.16
1.17	Exercise 1.17
1.18	Exercise 1.18
1.19	Exercise 1.19
1.20	Exercise 1.20
1.21	Exercise 1.21
1.22	Exercise 1.22
1.23	Exercise 1.23
1.24	Exercise 1.24
1.25	Exercise 1.25
1.26	Exercise 1.26
1.27	Exercise 1.27
1.28	Exercise 1.28
1.29	Exercise 1.29
1.30	Exercise 1.30
1.31	Exercise 1.31
1.32	Exercise 1.32
1.33	Exercise 1.33
1.34	Exercise 1.34
1.35	Exercise 1.35
1.36	Exercise 1.36
1.37	Exercise 1.37
1.38	Exercise 1.38
1.39	Exercise 1.39
1.40	Exercise 1.40
1.41	Exercise 1.41
1.42	Exercise 1.42
1.43	Exercise 1.43
1.44	Exercise 1.44
1.45	Exercise 1.45
1.46	Exercise 1.46

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1.34 Exercise 1.34

Suppose we have the following procedure f:

(define (f g)
(g 2))

If we ask the interpreter to evaluate (f f), the following happens:

(f f)

= (f 2)

= (2 2)

This fails, because 2 is not a procedure.

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