What is &&& operation in C

#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’

There is no `&&&` operator or token in C. But the `&&` (logical "and") and `&` (unary address-of or bitwise "and") operators do exist.

By the [maximal munch](

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) rule, this:

c = i &&& i;

is equivalent to this:

c = i && & i;

It sets `c` to 1 if both `i` and `&i` are true, and to 0 if either of them is false.

For an int, any non-zero value is true. For a pointer, any non-null value is true (and the address of an object is always non-null). So:

It sets `c` to 1 if `i` is non-zero, or to `0` if `i` is equal to zero.

Which implies that the `&&&` is being used here just for deliberate obfuscation. The assignment might as well be any of the following:

c = i && 1;
c = !!i;
c = (bool)i; // C++ or C with <stdbool.h>
c = i ? 1 : 0; /* C */
c = i ? true : false; // C++