Why is "else" a special symbol in scheme? Can it be defined as a procedure?
lisp, scheme
Solution
If it were an ordinary variable, you could do:
(set! else #f)
and then all the `cond` expressions that depend on the `else` clause being executed would stop working.
Problem
I started learning Scheme recently by reading SICP. In the opening chapter, it goes over conditionals and it talks about using `else` within the `cond` "special form" - which to my understanding is defined as "something the interpreter "just knows about". My question, is why is `else` defined as a "special form" and not as a procedure? If I fire up my mit-scheme interpreter, and type: `(else 1)` it raises an error. If I define something like `(define (myelse x) x)`, I can use it in the same way it is used within the `cond` expression like: ``` (define (abs x) (cond ((< x 0) (- x)) (myelse x))) ``` So why is `else` treated as something special, and not defined in scheme itself?