Why does if( -8 & 7) return false

Question

When i try to run the following code it prints "FALSE" instead of "TRUE" Can somebody explain why the code returns false?

#include <stdio.h>

int main(void)
{
    if(-8 & 7)
    {
        printf("TRUE");
    }
    else
    {
        printf("FALSE");
    }
    return 0;
}

Do you know the meaning of the `&` operator (as opposed to the `&&` operator)? Do you know the standard binary representation of the integer 7? What [common representations for negative numbers](https://en.wikipedia.org/wiki/Signed_number_representations) exist? What's common today? What's the binary representation of -8 in it? If you answer these basic questions you can answer your specific question here (and all others like it). — Peter - Reinstate Monica, Nov 24 '19 at 14:10

score 9 · Accepted Answer · answered Nov 24 '19 at 13:49

-8 can be represented the following way ( I will use a byte representation

8  = 00001000
~8 = 11110111
-8 = 11111000 (~8 + 1)

That is -8 in the two-complement representation is equal tp ~8 + 1

So -8 is equal to 11111000 and 7 is equal to 00000111

11111000
&
00000111
========
00000000

that is the binary AND operation yields a false result.

score 0 · Answer 2 · answered Nov 24 '19 at 14:21

0

7 is represented as 00000111 in binary. -8 is represented as 11111000 in binary. The bitwise AND operation performs an AND on every bit:

Hence, the if condition is false.

answered Nov 24 '19 at 14:21

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2 Answers2