Is the "else" keyword, used as final case in a cond-expression, a special form in Scheme?

Question

The cond-expression in Scheme is a special form, but is the else keyword, used as final case in a cond-expression, a special form? Or is it just a reserved keyword that is essentially equivalent to the truth-value #t ? In the latter case, why cannot I write something like (?eq else #t)?

Answer 1

The Scheme standards call else, as used in a cond form, auxiliary syntax. R6RS shows one possible implementation of cond using syntax-rules; here else is called a <literal>:

(define-syntax cond
  (syntax-rules (else =>)
    ((cond (else result1 result2 ...))
     (begin result1 result2 ...))
;; ...

Note that else is not a replacement for #t. A <literal> is an identifier that is used to match input subforms; it is treated as a syntactic keyword within the syntax-rules form.

Answer 2

It's part of the syntax of cond and case. R⁷RS specifies the following syntax:

(cond <cond clause>+)
(cond <cond clause>* (else <tail sequence>))

(case <expression>
  <case clause>+)
(case <expression>
  <case clause>*
  (else <tail sequence>))

It's not defined outside the syntax of these special forms.

Answer 3

Neither. You can test this by evaluating else outside of a cond: it behaves just like any other unbound symbol. The cond macro treats it specially, but there is nothing inherently special about it in any other context.