Chapter 1, Building Abstractions with Procedures
Section - Formulating Abstractions with Higher-Order Procedures
Exercise 1.41
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(define (double f)
(lambda (x) (f (f x)))
)
Now evaluating the statement given:
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> (((double (double double)) inc) 5)
The call double double returns a procedure, lets call this procedure X, that applies double twice to an argument passed to it.
And call to double (double double), return a procedure, lets call this procedure Y, that applies X twice to the argument passed to it.
And the invocation Y inc, returns a procedure, lets call it Z, that applies ‘X’ twice to inc. i.e. (X (X inc)).
Thus now finally the call to Z 5 is evaluated as follows:
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(Z 5)
((X (X inc)) 5)
((double (double (double (double inc))) 5))
Now we can see that in the innermost call, inc is called twice, than it is doubled thus called 4 times, again it is doubled thus called 8 times and again it is double thus called 16 times.
Thus thus function will return after calling inc 16 times starting with argument 5.
Thus the output is $16+5 = 21$.