How does std::cout print negative zero in a ones-complement system?

Question

On a ones-complement platform, what would the following code print?

#include 

int main() {
    int i = 1, j = -1;

    std::cout << i+         


        

Answer 1

Indeed adding n to -n should give you a negative zero. But the generation of -0 doesn't happen in practice since 1's complement addition uses a technique called a complementing subtractor (the second argument is complemented and subtracted from the first).

(The idiomatic way of getting a signed floating point zero doesn't apply here since you can't divide an integer by zero).