[erlang-questions] illegal guard expression for IF illegal guard expression for "if"
CGS
cgsmcmlxxv@REDACTED
Fri Dec 2 11:06:18 CET 2011
If statement is not a block statement (to fix the variables values at
the entrance of the statement) in Erlang. That means, if an
uniform:random() value (say, 0.4) is used for checking the first branch
condition and the value is changed within the branch (say, 0.7), then
you enter a paradox related to if the function should be considered for
the first or the second branch. To avoid this, Erlang requires an
assurance that the compared values do not change while processing the if
statement.
CGS
On 12/02/2011 10:56 AM, Barco You wrote:
> Hi CGS,
>
> I can't understand your statement --- " if in the first branch you use
> again random:uniform(), the second branch condition can report an
> inaccurate result." Could you please make it clearer? Thanks!
>
> Hi Others,
>
> Why functions with side effect can not be in the guard expressions?
>
>
> Thank you!
> Barco
>
> On Fri, Dec 2, 2011 at 5:50 PM, CGS <cgsmcmlxxv@REDACTED
> <mailto:cgsmcmlxxv@REDACTED>> wrote:
>
> Hi,
>
> To put in simple words for better understanding, you can have only
> constant variables withing the guard expression. That means, in
> your case, if in the first branch you use again random:uniform(),
> the second branch condition can report an inaccurate result.
>
> Alternatively, you can use case statement:
>
> case (random:uniform()<0.5) of
> true -> good;
> false -> bad
> end
>
> I hope this answer will help you.
>
> CGS
>
>
>
>
> On 12/02/2011 10:30 AM, Barco You wrote:
>> Why does the following expression got "illegal guard expression"
>> when compiling:
>> X = 0.5,
>> if
>> random:uniform() < X -> %error reported for this line
>> good;
>> true ->
>> bad
>> end.
>>
>> But if I change it to following expression, it's ok:
>> X = 0.5,
>> Ran = random:uniform(),
>> if
>> Ran < X ->
>> good;
>> true ->
>> bad
>> end.
>>
>> BRs,
>> Barco
>>
>>
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>
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