How printf("%c\n",~('C'*-1)) is computed in c?

Question

#include<stdio.h>
int main()
{
printf("%c\n",~('C'*-1));
return 0;
}

I have tried the above source code and executed without any warnings.

The output is B. I am excited how the above code is processed and what is the meaning for printf("%c\n",~('C'*-1))

Answer 1

In C, 'C' is an int, it's a small integer with a value of 67 (assuming ASCII). You can get each step from:

#include<stdio.h>
int main()
{
    printf("%d\n", 'C');            //67
    printf("%d\n", 'C' * -1);       //-67
    printf("%d\n", ~('C' * - 1));   //66
    printf("%c\n",~('C' * -1));     //B
    return 0;
}

In 2's complement, the value of ~(-67) is 66.

Answer 2

The only important part is this expression:

~('C' * -1)

Let's break it down:

• 'C' is ASCII code 67.
• ('C' * -1) is -67.
• -67 is, in binary, 10111101
• Bitwise negate that (with ~), and you have 01000010, which is 66.
• 66 is the ASCII code for 'B'.

More generally, most computers use "two's complement" arithmetic, where numerical negation followed by bitwise negation is equivalent to subtracting 1. And of course B is one less than C in ASCII.

On a computer that doesn't use two's complement arithmetic, the result may be different. Such computers are rare.