c/c++ left shift unsigned vs signed

Question

I have this code.

#include 

int main()
{
    unsigned long int i = 1U << 31;
    std::cout << i << std::endl;
    unsigned lon         


        

Answer 1

I think it is compiler dependent .
It gives same value
2147483648 2147483648
on my machiene (g++) .
Proof : http://ideone.com/cvYzxN

And if overflow is there , then because uwantsum is unsigned long int and unsigned values are ALWAYS positive , conversion is done from signed to unsigned by using (uwantsum)%2^64 .

Hope this helps !