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.