Scheme: Why is there the need to use a cond here?

Question

I tried to write a (simple, i.e. without eqan?) one? function like such:

(define one?
  (lambda (n)
    ((= 1 n))))

But the above doesn't work though because when I call it like such:

(one? 1)

I get greeted with this error:

procedure application: expected procedure, given: #t (no arguments)

---

The correct way (from The Little Schemer) to write it is:

(define one?
  (lambda (n)
    (cond
      (else (= 1 n)))))

Why is there a need to use a cond with an else clause, instead of just returning (= 1 n) ?

Answer 1

There isn't any reason why you would want to do that. I'll check my copy of TLS when I get home to see if I can divine what's going on, but you're not missing anything fundamental about cond or anything.

Response to your note above: It's not working because you have an extra set of parentheses in the body of the lambda. It should be

(lambda (n) (= 1 n))

The extra parentheses in your version mean that instead of returning the value #t or #f, you're trying to call that value as a function with no arguments.

Answer 2

not having a copy of The Little Schemer handy, your example looks as if should work. I think the cond is extraneous. In psudeo-C the equivant (with cond) is:

int
one(int n)
{     
    switch (foo) {
        default:
           return  1 == n;
    }
}