(define (ex1 v1 v2) (let ([f ...]) (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)]) (f v1) (g v2) (g 0))))so that the value of (ex2 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 (ex3 v1 v2) (let ([mk ...]) (let ([f (mk)] [g (mk)]) (f v1) (g v2) (g 0))))so that the value of (ex3 n m) is m for any integers n and m. As always, 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)]) (f v1 v2) (g v3 v4) (list (f 0 0) (g 0 0)))))so that the value of (ex4 n m p q) is _(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: Saturday, October 14th, 2000mflatt@cs.utah.edu |