[eeps] EEP XXX: Pattern-test operator

Richard O'Keefe <>
Tue Apr 24 04:06:18 CEST 2012


On 24/04/2012, at 10:52 AM, Robert Virding wrote:

> I have missed one part of the discussion here and that is about ?=:
> why use ?= in a guard and not =, and what does ?= mean outside a
> guard?

    Pat = Expr
    in an expression has the value and effect of
    case Expr of X = Pat -> X end.

    Combine this with the fact that in a guard, an expression is now
    allowed as a guard test.

    bar(X) when X -> 1;
    bar(X)        -> 2.

    2> foo:bar(true).
    1.
    3> foo:bar(false).
    2.
    4> foo:bar([]).
    2.

    The problem now is that EITHER
	f(X) when X = [] -> ...
        > f([])
    fails (because the match X = [] succeeds but has value [] which is
    taken as false in a guard) OR the behaviour of = in a guard is
    inconsistent with its behaviour in an expression.

    THEREFORE
	Pat ?= Expr
    is defined to have the value of
	case Expr of Pat -> true ; _ -> false end
    (er, this is just what it says about the _value_, ok?)
    and to have this value in ALL contexts, both guards and expressions.    
> 
> I think if we could avoid introducing a new operator just for this
> would be good.

We could have.  But once 'and' 'or' 'andalso' 'orelse' and so on were
allowed in guards, it was too late.  As far as I can see, we have just
three choices:

	(0) the status quo
	(1) introduce a new operator
	(2) give Pat = Expr the value of Expr in an expression
	    but true/false in a guard, which means that there
	    will be expressions that are legal in both guard and
	    expression contexts and will succeed quietly in both
	    but with different values.




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