[erlang-questions] illegal guard expression for IF illegal guard expression for "if"

Barco You barcojie@REDACTED
Fri Dec 2 15:42:46 CET 2011


Ooooh! Maybe I understand now!

It means that if the first branch is not satisfied, saying get 0.7, then
random:uniform() would be evaluated again in the second branch and it
possible to get a value less than 0.5, which then makes a paradox. Right?

In the contrast, with case statement, the condition is only evaluated once
no matter whichever branch would be satisfied.

How smart erlang is, and how amazing you erlangers are!

Barco
On Dec 2, 2011 6:18 PM, "CGS" <cgsmcmlxxv@REDACTED> wrote:

> **
> Just to make it clearer:
>
> The second condition is equivalent with the expression: "if the value of
> random:uniform() is greater or equal to 0.5 then report 'bad'". In this
> case, if you entered with value 0.4 in the first branch and modified it to
> 0.7 in the first branch, the second branch should be executed as well. That
> is against the if statement purpose which should report only one branch at
> the time as valid. Therefore, Erlang requires for the compared values to be
> constant during the if statement processing.
>
> CGS
>
>
>
> On 12/02/2011 11:06 AM, CGS wrote:
>
> 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> 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
>>
>>
>> _______________________________________________
>> erlang-questions mailing listerlang-questions@REDACTED://erlang.org/mailman/listinfo/erlang-questions
>>
>>
>>
>> _______________________________________________
>> erlang-questions mailing list
>> erlang-questions@REDACTED
>> http://erlang.org/mailman/listinfo/erlang-questions
>>
>>
>
>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://erlang.org/pipermail/erlang-questions/attachments/20111202/57b2cca7/attachment.htm>


More information about the erlang-questions mailing list