[erlang-questions] guard expression restriction
Fri Dec 3 02:20:16 CET 2010
On 2/12/2010, at 11:44 PM, Hynek Vychodil wrote:
> On Thu, Dec 2, 2010 at 9:37 AM, Kostis Sagonas <kostis@REDACTED> wrote:
>> Richard O'Keefe wrote:
>>> There is of course a fairly elementary transformation by means of
>>> which anyone who *really* needs a function call in a guard finds
>>> out that they don't.
>>> f(Arguments) when Expression -> Body;
>>> <rest of f>
>>> f(Arguments) ->
>>> case Expression
>>> of true -> Body
>>> ; false -> f'(Arguments)
>>> f(Vars) -> f'(Vars).
>>> where f' is the rest of f with the name changed.
> Anyway, this rewritten version doesn't behaves like guards
It isn't *MEANT* to. That's the point, really.
If you are calling user-defined functions you SHOULDN'T be
suppressing exceptions. (Haskell and Clean don't...)
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