C++ 隐式转换(有符号 + 无符号)

I understand that, regarding implicit conversions, if we have an unsigned type operand and a signed type operand, and the type of the unsigned operand is the same as (or larger) than the type of the signed operand, the signed operand will be converted to unsigned.

所以:

unsigned int u = 10;  
signed int s = -8;

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

//prints 2 because it will convert `s` to `unsigned int`, now `s` has the value
//4294967288, then it will add `u` to it, which is an out-of-range value, so,
//in my machine, `4294967298 % 4294967296 = 2`

我不明白 - 我读到如果有符号操作数的类型大于无符号操作数:

signed long long s = -8;
unsigned int u = 10;
std::cout << s + u << std::endl;

#include <iostream>

signed int s1 = -4;
unsigned int u1 = 2;

signed long int s2 = -4;
unsigned int u2 = 2;

signed long long int s3 = -4;
unsigned long int u3 = 2;

int main()
{
    std::cout << (s1 + u1) << "
"; // 4294967294
    std::cout << (s2 + u2) << "
"; // -2 
    std::cout << (s3 + u3) << "
"; // 18446744073709551614  
}

#include <iostream>

signed int s1 = -4;
unsigned int u1 = 2;
int main()
{
    std::cout << (s1 < u1 ? "s1 < u1" : "s1 !< u1") << "
";  // "s1 !< u1"
    std::cout << (-4 < 2u ? "-4 < 2u" : "-4 !< 2u") << "
";  // "-4 !< 2u"
}