Integer promotion unsigned in c++

Question

int main() {
    unsigned i = 5;
    int j = -10; 
    double d = i + j;
    long l = i + j;
    int k = i + j;
    std::cout << d << "\n";     //4.29497e+09
    std::cout << l << "\n";     //4294967291
    std::cout << k << "\n";     //-5
    std::cout << i + j << "\n"; //4294967291
}

I believe signed int is promoted to unsigned before doing the arithmetic operators.
While -10 is converted to unsigned unsigned integer underflow ( is this the correct term?? ) will occur and after addition it prints 4294967291 .

Why this is not happening in the case of int k which print -5 ?

Answer 1

The process of doing the arithmetic operator involves a conversion to make the two values have the same type. The name for this process is finding the common type , and for the case of int and unsigned int , the conversions are called usual arithmetic conversions . The term promotion is not used in this particular case.

In the case of i + j , the int is converted to unsigned int , by adding UINT_MAX + 1 to it. So the result of i + j is UINT_MAX - 4 , which on your system is 4294967291 .

You then store this value in various data types; the only output that needs further explanation is k . The value UINT_MAX - 4 cannot fit in int . This is called out-of-range assignment and the resulting value is implementation-defined. On your system it apparently assigns the int value which has the same representation as the unsigned int value.

Answer 2

j will be converted to unsigned int before addition, and this happens in all your i + j . A quick experiment.

In the case of int k = i + j . As in the case of your implementation and mine, i + j produces: 4294967291 . 4294967291 is larger than std::numeric_limits<int>::max() , the behavior is going to be implementation defined. Why not try assigning 4294967291 to an int ?

#include <iostream>

int main(){
    int k = 4294967291;
    std::cout << k << std::endl;
}

Produces:

-5

As seen Here