[erlang-questions] Why I cannot use functions in gaurds

Richard O'Keefe ok@REDACTED
Wed Nov 7 04:26:04 CET 2012

On 6/11/2012, at 7:57 PM, Maruthavanan Subbarayan wrote:

> I would like to know  if the first one is valid why not the second one?

(a) Because the language definition has always said so.
(b) This topic has been beaten to death in this mailing list
    and is explained in every Erlang book I've seen.
(c) The short answer is that GUARDS ARE NOT EXPRESSIONS.
(d) The long answer traces this back to concurrent logic programming
    languages (look at Flat Concurrent Prolog and Strand88) and to
    the fact that guards are supposed to be certain to terminate in
    a finite time AND to have no side effects.
(e) If/when Erlang gets abstract patterns, it will be as safe to
    use them in guards as anything else currently allowed in guards.

It is _always_ possible to reshape your design so there is no
use of functions in guards, and the result is _usually_ clearer.
> case L of
>    {answer, N} when N == M:F(A) -> true;
>    _ -> false
>  end.

In this case you can write

	X = M:F(A),
	case L
	  of {answer, X} -> true
	   ; _           -> false
	case L
	  of {answer, X} ->
	     case M:F(A)
	       of X -> true
	        ; _ -> false
	   ; _      -> false

More information about the erlang-questions mailing list