Question

I am asked to interpret the output of a function with specific inputs but I don’t understand how the function works. It is suppose to be a new version of if but to me it looks like it does nothing at all.

(define (if-2 a b c)
    (cond (a b)
    (else c)))

To me this just looks like it will always print b but I am not sure.

Answer 1

It seems like you are unfamiliar with the cond form. It works like this:

(cond
  ((<predicate1> <args>) <actions>)
    ;^^-- this form evaluates to true or false. 
    ;  If true, we do <actions>, if not we move on to the next predicate.
  ((<predicate2> <args>) <actions>) ; We can have however many predicates we wish
  (else ;<-- else is always true.  This is our catch-all.
    <actions>))

Below is your code with some variables renamed.

(define (if-2 predicate arg1 arg2)
    (cond
      (predicate arg1)
      (else arg2)))

To figure out why it is always returning arg1 for your tests, recall that Scheme sees everything as true except the explicit false symbol (usually #f) and the empty list '().

So when you call (if-2 > 2 3) the cond form evaluates to this:

(cond
  (> 2)
  ;^---- `>` is not the empty list, so it evals to #t 
  (else 3))                              

Then since cond returns the first thing it finds to be associated with a true value you get 2 back.

To make if-2 work as expected you need to call it differently, e.g. (if-2 (> 3 2) 'yep! 'nope!) will return 'yep! since 3 is greater than 2.