Hi,
Reply embedded.
--- In [email protected], "Shyan Lam" <[EMAIL PROTECTED]> wrote:
> > -----Original Message-----
> > From: [email protected] [mailto:[EMAIL PROTECTED] On Behalf
> > Of Pedro Izecksohn
> > Sent: Saturday, May 26, 2007 10:15 PM
> > If any value not zero is true why "true which expands to the integer
> > constant 1"?
> >
> > I wasted a week to find a bug that I wrote like:
> >
> > if ((x&y)==true) //etc
>
> '&' is a bitwise-AND operator. I think what you really meant is
> "logical-AND" (&&).
>
> if ((x && y) == true)
>
> Unless you want 'y' to be evaluated regardless of the result of 'x'.
>
>
> > but I should have written just:
> >
> > if (x&y) //etc
>
> It still can be shorten to:
>
> if (x && y)
Why can it be shortened?
are you sure x&y == x&&y???
#include <stdio.h>
int main()
{
int x = 2, y =4;
printf(" bit AND == %d, logical AND == %d\n", (x & y),(x && y) );
return 1;
}
Output: bit AND == 0, logical AND == 1
Am I missing something?
Regards,
-Saurabh
