Expert: Narendra - 10/31/2004

Question
respected sir namaskar,
sir if i write the following;

{
int i=2, j=-1, k=3, a;
a= ++i || ++j && ++k;
printf ("%d  %d  %d",i, j, k);
}
i get the answer :
i=3 j=-1  and k=3


sir, as a general rule && is executed before the evaluation of || so the answer should have been

i=2  j=0  and k =3
but it is not happening . Why the precedence rule is not being followed here.

thanks for listening.  

Answer
Here is how it happens:
Depending on the order of precedence, your expression will be grouped like this:
a = ++i || (++j && ++k);
Here there are two expressions and will be evaluated from left to right.
Since ++i tests positive, the rightmost expression will never get executed!
So, you end up with values i=3, j=-1 and k=3.

If you just want to test the operator precedence it is ok.
But, if you want to do serious programming, then use parentheses and you will never get into all this confusion.

-ssnkumar