[erlang-questions] guard expression restriction

Hynek Vychodil hynek@REDACTED
Thu Dec 2 11:44:32 CET 2010

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)
>>            end;
>>        f(Vars) -> f'(Vars).
>> where f' is the rest of f with the name changed.
> Yes, Erlang is Turing complete as it is (so you can express in it any
> algorithm you want to) and of course there is that transformation, but do
> you really claim that the bulky code with the case statement and having to
> factor the rest of f into a separate function is more expressive and
> programmer-friendly than the one liner with the guard? Sorry, but I do not
> see this.

Anyway, this rewritten version doesn't behaves like guards because it
should be more like:

        f(Arguments) ->
            case catch Expression
              of true -> Body
               ; _ -> f'(Arguments)
        f(Vars) -> f'(Vars).

or even worse

        f(Arguments) ->
            try Expression
              of true -> Body
               ; _ -> f'(Arguments)
                 _:_ -> f'(Arguments)
        f(Vars) -> f'(Vars).

because soon or late one would write throw(true) inside Expression.
>> I'm not providing an example, because I've tried four and each of
>> them was immediately better without trying to do this in guards...
> Hmmm... In any case, the paper I mentioned in my previous mail contains an
> example of more clear and succinct code when user-defined guards are
> allowed.
> Kostis
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--Hynek (Pichi) Vychodil

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