[erlang-questions] guard clause oddity, maybe bug (in my brain)?

[Jim]

Thank you both, I guess I don't  fully understand the associative and precedence of these operators. Having said that it is surprising to me that  
(not is_tuple(Elt)) andalso ?IS_INDEX(I)
can ever evaluate to true when Elt is a tuple.  Andalso short circuits so it's curious to me that a fully parenthesized clause that evaluates to false to the left of the andalso wouldn't short circuit everything regardless of what happens to the right of the andalso.

> On Nov 20, 2015, at 3:46 PM, Pierre Fenoll <pierrefenoll@REDACTED> wrote:
> 
> Dmitry is on point. 
> 
> 1. Remove the useless braces around not is_tuple...
> 2. Add braces around ?IS_INDEX...
> 
> If that fixes it, think about moving the braces inside the macro's definition. 
> 
>> On 20 Nov 2015, at 16:03, Dmitry Kolesnikov <dmkolesnikov@REDACTED> wrote:
>> 
>> Hello,
>> 
>> How does you macro IS_INDEX looks like?
>> e.g.
>> 
>> 5> A = {1}.
>> {1}
>> 6> (not is_tuple(A)) andalso is_number(A) orelse A > 1.
>> true
>> 
>> but
>> 
>> 8> (not is_tuple(A)) andalso (is_number(A) orelse A > 1).
>> false
>> 
>> 
>> Best Regards, 
>> Dmitry
>> 
>> 
>>> On Nov 20, 2015, at 4:03 PM, jim rosenblum <jim.rosenblum@REDACTED> wrote:
>>> 
>>> I have a function:
>>> 
>>> walk ([I|Is], [Elt|_]=Ts) when (not is_tuple(Elt)) andalso ?IS_INDEX(I) ->
>>>  Element = index_to_element(I,Ts),
>>>  case 
>>>      ... snip ...
>>>  end.
>>> 
>>> This function clause gets executed even though Elt is a tuple. 
>>> 
>>> I have a break point within this function clause. I am looking at the 
>>> debugger at the break-point within the clause, and I can plainly see that Elt
>>> IS a tuple: within the debugger I can evaluate is_tuple(Elt) and it returns true,
>>> I can evaluate (not is_tuple(Elt)) and correctly get FALSE.
>>> 
>>> So.... how can I be in this clause when the guard evaluates to false?
>>> 
>>> Thoughts?
>>> 
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