(define (ex1 v1 v2) (let ([f ...]) (begin (f v1) (f v2))))so that the value of (ex1 n m) is n*m for any integers n and m. The catch is that neither v1 nor v2 may appear in anywhere within the expression replacing "...".
(define (ex2 v1 v2) (let ([mk ...]) (let ([f (mk)] [g (mk)]) (begin (g v1) (f v2) (g 1)))))so that the value of (ex2 n m) is n for any integers n and m. As always, neither v1 nor v2 may appear in anywhere within the expression replacing "...".
(define (ex3 v1 v2) (let ([mk ...]) (let ([f (mk)] [g (mk)]) (begin (f v1) (g v2) (g 1)))))so that the value of (ex3 n m) is n*m for any integers n and m. As before, neither v1 nor v2 may appear in anywhere within the expression replacing "...".
(define (ex4 v1 v2 v3 v4) (let ([mk ...]) (let ([f (mk)] [g (mk)]) (begin (f v1 v2) (g v3 v4) (list (f 1 1) (g 1 1))))))so that the value of (ex4 n m p q) is (list n*m*q m*p*q) for any integers n, m, p, and q. Of course, v1, v2, v3, or v4 may not appear in anywhere within the expression replacing "...".
Last update: Monday, October 8th, 2001mflatt@cs.utah.edu |