#include <stdio.h>
volatile int i;
int main()
{
int c;
for (i = 0; i < 3; i++)
{
c = i &&& i;
printf("%d\n", c);
}
return 0;
}
The output of the above program compiled using gcc
is
0
1
1
With the -Wall
or -Waddress
option, gcc
issues a warning:
warning: the address of ‘i’ will always evaluate as ‘true’ [-Waddress]
How is c
being evaluated in the above program?
It's c = i && (&i);
, with the second part being redundant, since &i
will never evaluate to false
.
For a user-defined type, where you can actually overload unary operator &
, it might be different, but it's still a very bad idea.
If you turn on warnings, you'll get something like:
warning: the address of ‘i’ will always evaluate as ‘true’