Why `-1 + unsigned(0) > 0` in C++?

Question

The following code output -1 1 in C++.

If i is evaluated as -1, why is it larger than 0?

#include <stdio.h>

int main() {
  auto i = -1 + unsigned(0);
  printf("%d %d\n", i, i > 0);
}

Answer 1

If i is evaluated as -1

Thats a wrong premise.

i is not -1. Adding 0u (or unsigned(0)) to -1 results in an unsigned. You may see -1 as output for i because using the wrong format specifier with printf is undefined.

This

#include <iostream>

int main() {
  auto i = -1 + unsigned(0);
  std::cout << i << " " << (i>0);
}

does not rely on guessing the type of i and prints:

4294967295 1

unsigned(-1) is the largest unsigned integer representable as unsigned.