Question

I am just trying to randomly make a list and use it in a larger function.

        (define make-random-list
        (if
          (= (random 2) 0) (list 2 3)
                           (list 3 2)))

This only produces the list (2 3) and I am not sure why. What is happening to cause this?

I can make the function work if I write it like this

  (define make-random-list
   (lambda (x)
   (if
    (= (random x) 0) (list 2 3)
                     (list 3 2))))

and calling (make-random-list 2)

but I do not understand why that would work and the other one would not. What is going on with scheme that would not allow the first function to produce random results?

Answer 1

In your first snippet, you’re assigning the result of a one-time computation to a variable. What you need here is to define a function that will be evaluated every time it’s invoked. Your second snippet does exactly that, but there is a shorter way to express the same:

(define (make-random-list x)
(if
  (= (random x) 0) (list 2 3)
                   (list 3 2)))

Note the difference in the syntax: a function definition encloses the function definition together with the formal argument names in parentheses, while there are no parentheses around the name of a variable.