Re: Unpredictable nature of increment operators

Martin York wrote:
> On May 8, 5:16 pm, bintom <binoythomas1...@[EMAIL PROTECTED]
> wrote:
>> Why doe the following C++ code behaves as it does?
>>
>> int i;
>>
>> i=1;            cout << (++i)++;          Output: 2
>> i=1;            cout << ++(++i);          Output: 3
>> i=1;            cout << (i++)++;          Output: Error. LValue
>> required i=1;            cout << ++(i++);          Output: Error.
>> LValue required
>
> Because code like that is so hard to read you would never actually
> write it!
>
> Also remember that it is illegal to modify a variable more than once
> within the same expression [undefined behavior I believe](somebody
> with a copy of the standard will be able to quote you chapter an verse
> and the actual definition rather than my layman's wording).

Only a variable of a built-in type, and only if there is no sequence
point between the modifications.  For example, it's not undefined
behaviour to do

    f(i++) + ++i;

IIRC, but the results are unspecified.  And if your class has the
operators ++() or ++(int) overloaded, the results are even well-
defined and predictable.  For 'int' they aren't...

V
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