Scheme problem (using a function as a parameter)

Question

I'm a Scheme newbie and trying to make sense of my homework. I've a function I made earlier called duplicate, and it looks like this:

( DEFINE ( duplicate lis )
          (IF (NULL? lis) '())
          ((CONS (CAR lis) (CONS (CAR lis) (duplicate (CDR lis))))
         ))

A typical i/o from this would be i: (duplicate '(1 2 3 4)) o: (1 1 2 2 3 3 4 4), so basicly it duplicates everything in the list. Moving on: Now I'm supposed to make a function that's called comp. It's supposed to be built like this:

(DEFINE (comp f g) (lambda (x) (f (g (x))))

Where I could input '(1 2 3 4) and it would return (1 1 4 4 9 9 16 16)

so f = duplicate and g = lambda. I know lambda should probably look like this:

(lambda (x) (* x x))

But here's where the problem starts, I've already spent several hours on this, and as you can see not made much progress.

Any help would be appreciated. Best regards.

1. In comp() you can't have (x); xis not a function: it should be (g x) instead of (g(x)) 2. To use comp with lambda , your (lambda(x..)) needs to work on a list to be compatible with double; as it is, this lambda only works on a single number. That is hard to do without loop. — frayser
– frayser, Commented Nov 15, 2010 at 14:04

Vijay Mathew · Accepted Answer · 2010-11-15 12:40:20Z

2

Use map:

> (map (lambda (x) (* x x)) (duplicate '(1 2 3 4)))
=> (1 1 4 4 9 9 16 16)

or, modify duplicate to take a procedure as its second argument and apply it to each element of the list:

(define (duplicate lst p)
  (if (null? lst) ()
      (append (list (p (car lst)) (p (car lst))) (duplicate (cdr lst) p))))

> (duplicate '(1 2 3 4) (lambda (x) (* x x)))
=> (1 1 4 4 9 9 16 16)

edited Nov 15, 2010 at 12:40

answered Nov 15, 2010 at 11:44

Vijay Mathew

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Comments

user445161 · Accepted Answer · 2010-11-15 16:53:29Z

0

One way to do is as follows:

(define (comp f g) (lambda (x) (f (g x))))
(define (square x) (* x x))
(define (dup x) (list x x))
(define (duplicate-square lst)
  (foldr append '() (map (comp dup square) lst)))

Now at the repl, do:

> (duplicate-square '(1 2 3 4))
'(1 1 4 4 9 9 16 16)

answered Nov 15, 2010 at 16:53

user445161

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