scheme - equal function with lists

I wrote this program:

(define find-combination
  {lambda (a b)
    (if (eq? ((quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b)))))
         (display "1*v1" + ((quotient (car a) (car b))*"v2"))
         (display "0*v1" + "0*v2"))})

(find-combination (list 2 2) (list 2 1))

a and b are two lists. Its give me the next problem: proc开发者_C百科edure application: expected procedure, given: 1; arguments were: 2.

I didn't get what is the problem. Someone can help me? Thank u.

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First of all you have too much brackets after eq? - what you wrote means evaluating (quotient (car a) (car b)) and treating it as a function with argument (quotient (car (cdr a)) (car (cdr b))). The error means that first thing was evaluated to 1 and your interpreter expected it to be a procedure, not an integer. This line should be:

(if (eq? (quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b))))

or even:

(if (eq? (quotient (car a) (car b)) (quotient (cadr a) (cadr b)))

Apart from that, lines with display calls are wrong - Scheme doesn't have an infix notation, so + and * are out of place.

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First of all you have a set of curly braces in your code(the one before lambda)

Also you have another set of paranthesis around the parameters you passed to eq? It should be something like this:

(eq? (quotient (car a) (car b)) (quotient (car (cdr a)) (car (cdr b))))

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In Scheme and Racket, parentheses change the meaning of things.

1

is a number, but

(1)

is a call to 1 as a function... but 1 is a number, not a function, so this will cause the error you describe.

Your use of curly braces is also a little unsettling to me.