Re: Cut and the operators mu and nu Joo ChurlSoo 06 Apr 2006 10:32 UTC

 * From: Jens Axel SÞgaard <xxxxxx@soegaard.net>
 | The specifications of mu and nu are:
   (mu <expr> ...)          => (lambda (f) (f <expr> ...))
   (nu <expr> ... <exprn>)  => (lambda (f) (apply f <expr> ... <exprn>))
   The <exprn> should be a list.

 | These are indeed useful operations. So useful that we already have
 | a srfi for them, namely srfi-26.
 | (cut <> <expr> ...) => (lambda (f) (f <expr> ...))
 | (cut apply <> <expr> ... <exprn>) => (lambda (f) (apply <expr> ... <exprn>))

It matters little whether we use (lambda (f) (f 1 2 3)) or (cut <> 1 2 3) or
(mu 1 2 3) or any other its equivalent.  The point is we know they can be used
in this way.

--
Nothing is complete unless you put it in final shape.