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

The literal 1 with no U is a signed int, so when you shift << 31, you get integer overflow, generating a negative number (under the umbrella of undefined behavior).

Assigning this negative number to an unsigned long causes sign extension, because long has more bits than int, and it translates the negative number into a large positive number by taking its modulus with 2⁶⁴, which is the rule for signed-to-unsigned conversion.