Scheme: About cond

Question

(cond ((test-1) (expression-1)))

When i use a cond, can i give the several functions in (expression-1)?

Like this:

(cond ((= 1 1) ((fun1) (fun2)) )

Answer 1

The begin is actually optional -- cond (in Scheme as well as in Emacs Lisp, at least) take multiple expressions after each test expression and evaluate them in turn in an implicit begin

(cond ((= 1 1) (fun1 ...) (fun2 ...))
      (t (something-else)))

Use #t instead of t for Scheme