Why is cond a special form in Scheme, rather than a function?

Question

(defun triangle-using-cond (number)
  (cond 
    ((<= number 0) 0) ; 1st
    ((= number 1) 1)  ; 2nd
    ((> number 1)     ; 3rd
      ;; 4th
      (+ number
         (triangle-using-cond (1- number))))))

Things that I know about Cond

• It allows multiple test and alternative expressions
• It has pre-specified evaluation order. For instance, the first condition will always evaluated whether it is right or not

One thing that I cannot distinguish is that what makes cond different from a function!

Answer 1

A function call (e0 e1 e2) is evaluated like this

1. e0 is evaluated, the result is (hopefully) a function f
2. e1 is evaluated, the result is a value v1
3. e2 is evaluated, the result is a value v2
4. The function body of `f` is evaluated in an environment in which
   the formal parameters are bound to the values `v1` and `v2`.

Note that all expressions e0, e1, and, e2 are evaluated before the body of the function is activated.

This means that a function call like (foo #t 2 (/ 3 0)) will result in an error when (/ 3 0) is evaluated - before control is handed over to the body of foo.

Now consider the special form if. In (if #t 2 (/ 3 0)) the expressions #t is evaluated and since the value non-false, the second expression 2 is evaluated and the resulting value is 2. Here (/ 3 0) is never evaluated.

If in contrast if were a function, then the expressions #t, 2, and, (/ 3 0) are evaluated before the body of is activated. And now (/ 3 0) will produce an error - even though the value of that expressions is not needed.

In short: Function calls will always evaluate all arguments, before the control is passed to the function body. If some expressions are not to be evaluated a special form is needed.

Here if and cond are examples of forms, that don't evaluate all subexpressions - so they they need to be special forms.