Why prefer signed over unsigned in C++? [closed]

Answer 1

It gives unexpected result when doing simple arithmetic operation:

unsigned int i;
i = 1 - 2;
//i is now 4294967295 on a 64bit machine

It gives unexpected result when doing simple comparison:

unsigned int j = 1;
std::cout << (j>-1) << std::endl;
//output 0 as false but 1 is greater than -1

This is because when doing the operations above, the signed ints are converted to unsigned, and it overflows and goes to a really big number.