Question

I have found this question about the special function “or” in scheme:

Joe Hacker states loudly that there is no reason or in Scheme needs to be special — it can just be defined by the programmer, like this:

(define (or x y)
    (if x
        #t
        y))

Is Joe right?

I can’t figure out why it shouldn’t be possible to do that.

Could some scheme-expert please explain if this works, and if no: why not?

Answer 1

It’s because this version of or evaluates all of its arguments (since it’s a function), while the standard Scheme or (which is not a function but special syntax) doesn’t. Try running (or #t (exit)) at the Scheme REPL and then try the same with your or function.

The behavior of the standard or is sometimes called short-circuited: it evaluates only those arguments that it needs to. This is very common for the binary boolean operator (or and and) across programming languages. The fact that or looks like a function call is a feature of Scheme/Lisp syntax, but looks deceive.