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lispschemetail-recursionsicp

Deconstructing a recursive process - SICP

Consider the following definition:

   (define foo
     (lambda (x y)
        (if (= x y)
            0
            (+ x (foo (+ x 1) y)))))

What is the test expression? (write the actual expression, not its value)

I would think it is just (if (= x y) but the MIT 6.001 On Line Tutor is not accepting that answer.

Solution

  • The test would be:

    (= x y)

    That's the expression that actually returns a boolean value, and the behaviour of the if conditional expression depends on it - if it's #t (or in general: any non-false value) the consequent part will be executed: 0. Only if it's #f the alternative part will be executed: (+ x (foo (+ x 1) y)).