Checking for membership in an Erlang guard

Question

What is the simplest way to write an if statement in Erlang, where a part of the guard is member(E, L), i.e., testing if E is a member of the list L? The naive approach is:

if 
  ... andalso member(E,L) -> ...
end

But is does not work becuase, if I understand correctly, member is not a guard expression. Which way will work?

score 13 · Accepted Answer · answered Aug 03 '11 at 14:13

13

Member functionality is, as you say, not a valid guard. Instead you might consider using a case pattern? It's possibly to include your other if-clauses in the case expression.

case {member(E,L),Expr} of
  {true,true} -> do(), is_member;
  {true,false} -> is_member;
  {false,_} -> no_member
end

answered Aug 03 '11 at 14:13

D.Nibon

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Nice. I hoped to avoid using booleans as case clauses, but at least it spares me the need to nest a case statement inside an if statement. – Little Bobby Tables Aug 04 '11 at 08:03
1

Using booleans in case statements is quite common, probably more common than any use of "if". – YOUR ARGUMENT IS VALID Aug 04 '11 at 13:54

score 6 · Answer 2 · answered Aug 03 '11 at 14:11

6

It is not possible to test list membership in a guard in Erlang. You have to do this:

f(E, L) ->
    case lists:member(E, L) of
        true  -> ...;
        false -> ...
    end.

answered Aug 03 '11 at 14:11

Adam Lindberg

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score 2 · Answer 3 · answered Aug 07 '11 at 15:43

The easiest thing is to consider guards as a part of pattern matching, the part which cannot, or is difficult to, express in the pattern itself. So a guard is a sequence of guard tests and not boolean expressions. The original guard syntax made it easier to see the difference but now they look like boolean expressions, which they are not.

Checking for membership in an Erlang guard

3 Answers3

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