Why is the output -33 for this code snippet

Question

#include

int main()
{
    int a=32;
    printf(\"%d\\n\", ~a);  //line 2
    return 0;
}

o/p = -33

Actually in the original snippet li

Answer 1

In your C implementation, as in most modern implementations of any programming language, signed integers are represented with two’s complement.

In two’s complement, the high bit indicates a negative number, and the values are encoded as in these samples:

Bits  Decimal
0…011 +3
0…010 +2
0…001 +1
0…000  0
1…111 -1
1…110 -2
1…101 -3

Thus, if the usual (unsigned) binary value for the bits is n and the high bit is zero, the represented value is +n. However, if the high bit is one, then the represented value is n-2^w, where w is the width (the number of bits in the format).

So, in an unsigned 32-bit format, 32 one bits would normally be 4,294,967,295. In a two’s complement 32-bit format, 32 one bits is 4,294,967,295 - 2³² = -1.

In your case, the bits you have are 1111 1111 1111 1111 1111 1111 1101 1111. In unsigned 32-bit format, that is 4,294,967,263. In two’s complement, it is 4,294,967,263 - 2³² = -33.