Le 20/04/2012 03:56, Darren New a écrit :
> On 4/19/2012 18:38, Alain wrote:
>> Le 2012/04/17 09:40, Warp a écrit :
>>
>>>> is a perfectly valid statement in C++. On the other hand, it also
>>>> means that
>>>
>>>> while (x[i] = y[i--]) ;
>>>
>>>> is perfectly valid. You sick, sick people.
>>>
>>> I think that's Undefined Behavior because the same variable is being
>>> modified and referenced more than once in the same expression.
>>>
>>
>> Nothing undefined here.
> 
> Does it compare x[i] to y[i] or to y[i-1]?  Why do you answer that way?
> 
You missed the point: let's name k the initial value of i;

Would you expect the test to be between:
 x[k] & y[k]
 x[k-1] & y[k]

It cannot be x[k] & y[k-1], per the definition of the post-decrement
operator.

-- 
A good Manager will take you
through the forest, no mater what.
A Leader will take time to climb on a
Tree and say 'This is the wrong forest'.

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