After coming across something similar in a co-worker’s code, I’m having trouble understanding why/how this code executes without compiler warnings or errors.
#include <iostream>
int main (void)
{
unsigned int u = 42;
const int& s = u;
std::cout << "u=" << u << " s=" << s << "\n";
u = 6 * 9;
std::cout << "u=" << u << " s=" << s << "\n";
}
Output:
u=42 s=42
u=54 s=42
First, I expect the compiler to issue some kind of diagnostic when I mix signed/unsigned integers like this. Certainly it does if I attempt to compare with <. That’s one thing that confuses me.
Second, I’m not sure how the second line of output is generated. I expected the value of s to be 54. How does this work? Is the compiler creating an anonymous, automatic signed integer variable, assigning the value of u, and pointing the reference s at that value? Or is it doing something else, like changing s from a reference to a plain integer variable?
>Solution :
References can’t bind to objects with different type directly. Given const int& s = u;, u is implicitly converted to int firstly, which is a temporary and then s binds to the temporary int. (Lvalue-references to const (and rvalue-references) could bind to temporaries.) The lifetime of the temporary is prolonged to the lifetime of s, i.e. it’ll be destroyed when get out of main.