I have a simple function that will find the two's Complement for an unsigned integer and then test it to make sure that it is correct

unsigned twoscomplement(unsigned v) {
};

int main()
{
    unsigned a = 255;
    unsigned c = twoscomplement(a);
    unsigned s = a c;
    printf("%u %u=%u\n", a, c, s);
    return 0;
} 

When I asked about how I would go around solving this I got the answer unsigned c = (~a) 1; From what I understood (~a) flips the bits and then 1 is for the overflow? Any help on this matter would be appreciated

CodePudding user response：

Whenever we work with one’s complement or two’s complement, we need to state what the word size is. If there are w bits in the word, then the one’s complement of a number x is obtained by subtracting x from the binary numeral made of w 1s. If w is 16, we use 1111111111111111₂, which is 65535. Then the one’s complement of x is 1111111111111111₂−x. Viewing x as a binary numeral (with at most w bits), whatever bits are on in x will be off in 1111111111111111₂−x, and whatever bits are off in x will be on in 1111111111111111₂−x. Hence, all the bits are complemented.

C has an operator for the one’s complement; ~x flips all the bits, so it produces the one’s complement of x.

The two’s complement of x is 2^w−x, by definition (except that the two’s complement of 0 is 0). 2^w equals one plus that binary numeral made of w 1s. For example, 2¹⁶ = 65535 1. Therefore, the two’s complement is one more than the one’s complement. Therefore the two’s complement of x is ~x 1.

C also has an operator for the two’s complement, for unsigned integers. Unsigned arithmetic is defined to “wrap” modulo 2^w; whenever a regular arithmetic result is outside that range, it is brought back into that range by adding or subtracting 2^w as needed. The regular arithmetic negation of x would be negative (if x is not zero), so the computed result of -x is −x 2^w = 2^w−x, which is the two’s complement of x.