How to run multiple expressions when an if condition is true?

Question

I'm new to Scheme and am trying to have an if-statement perform more than one action if its condition is true. I tried something like:

(if (char=? (string-ref str loc) #\a)
        ((+ count 1) (+ reference 1))
        ~else action, etc.~

And it complains about my action, saying

application: not a procedure

If I remove the parentheses, so that the action for a true condition is:

(+ count 1) (+ reference 1)

It complains

if: bad syntax

and fails to run at all. What am I missing?

Answer 1

There are two problems with the code. The first, an if form can't have more than one expression in the consequent and in the alternative. Two possible solutions for that - you can either use begin (not just a couple of parenthesis, that's for procedure application) to surround multiple expression:

(if (char=? (string-ref str loc) #\a)
    ; needs an explicit `begin` for more than one expression
    (begin
      (+ count 1)
      (+ reference 1))
    ; needs an explicit `begin` for more than one expression
    (begin
      ~else action, etc~))

... or use a cond, which is a better alternative because it already includes an implicit begin:

(cond ((char=? (string-ref str loc) #\a)
       ; already includes an implicit `begin`
       (+ count 1)
       (+ reference 1))
      (else
       ; already includes an implicit `begin`
        ~else action, etc~))

The second problem is more subtle and serious, both of the expressions in the consequent part are probably not doing what you expect. This one: (+ count 1) is doing nothing at all, the incremented value is lost because you didn't use it after incrementing it. Same thing with the other one: (+ reference 1), but at least here the value is being returned as the result of the conditional expression. You should either pass both incremented values along to a procedure (probably as part of a recursion):

(cond ((char=? (string-ref str loc) #\a)
       ; let's say the procedure is called `loop`
       (loop (+ count 1) (+ reference 1)))
      (else
        ~else action, etc~))

Or directly update in the variables the result of the increments, although this is not the idiomatic way to write a solution in Scheme (it looks like a solution in C, Java, etc.):

(cond ((char=? (string-ref str loc) #\a)
       ; here we are mutating in-place the value of the variables
       (set! count (+ count 1))
       (set! reference (+ reference 1)))
      (else
        ~else action, etc~))