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

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);
}

Use std::cout, then you won't risk guessing types wrong for format specifiers. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Apr 5, 2023 at 10:18
"If i is evaluated as -1": It is not. Your printf is just wrong. Avoid using it when writing C++. There is std::cout << instead and since C++20 also std::format. First ask yourself what type auto in auto i deduces to. You can verify your guess by using static_assert(std::is_same_v<decltype(i), /*guess*/>);. — user17732522
– user17732522, Commented Apr 5, 2023 at 10:21
Per the first answer: "In your case, we have one unsigned int and signed int. Referring to (3) above, since both operands have the same rank, [...] will need to be converted an unsigned integer." — E_net4
– E_net4, Commented Apr 5, 2023 at 10:25

463035818_is_not_an_ai · Accepted Answer · 2023-04-05 10:22:43Z

3

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.

answered Apr 5, 2023 at 10:22

463035818_is_not_an_ai

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