Re: Unpredictable nature of increment operators

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).