C++ unsigned and signed conversion

Question

I've seen this kind of question asked before, but supplied answers don't clear everything for me. When this question is posted it is usually accompanied by next example:

#include <iostream>

int main()
{
    unsigned int u = 10;
             int i = -42;

    std::cout << i + i << std::endl;
    std::cout << i + u << std::endl;

    return 0;
}

Output:

-84
4294967264

All works as expected int is converted to unsigned. But if absolute value of i is less than u it seems like no such conversion is hapenning.

#include <iostream>

int main()
{
    unsigned int u = 10;
             int i = -3;

    std::cout << i + i << std::endl;
    std::cout << i + u << std::endl;

    return 0;
}

Output:

-6
7

I haven't seen any answer mentioning it and can't find any explanation. Even though it seems like a logical thing to happen I havent thing any explanation accounting for this.

Answer 1

After:

unsigned int u = 10;
         int i = -3;

the evaluation of i + u proceeds by first converting i to unsigned int. For a 32-bit unsigned int, this conversion wraps modulo 2³², which is 4,294,967,296. The result of this wrapping is −3 + 4,294,967,296 = 4,294,967,293.

After the conversion, we are adding 4,294,967,293 (converted i) and 10 (u). This would be 4,294,967,303. Since this exceeds the range of a 32-bit unsigned int, it is wrapped modulo 4,294,967,296. The result of this is 4,294,967,303 − 4,294,967,296 = 7.

Thus “7” is printed.