(define (ex1 v1 v2)
(let ([f ...])
(begin
(f v1)
(f v2))))so that the value of (ex1 n m) is (list m n) 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
(f (list v1))
(g (list v2))
(g '())))))so that the value of (ex2 n m) is (list 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)])
(begin
(g (list v1))
(f (list v2))
(g '())))))so that the value of (ex3 n m) is (list n) 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)])
(begin
(f (list v1) (list v2))
(g (list v3) (list v4))
(list (f '() '()) (g '() '()))))))so that the value of (ex4 n m p q) is (list (list n m q) (list m q p)) 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: Tuesday, October 8th, 2002mflatt@cs.utah.edu |