Is this a valid function?

Question

What happens to the reference in function parameter, if it gets destroyed when the function returns, then how const int *i is still a valid pointer?

const int* func(const int &x = 5)
{
    return &x;
}


int main()
{
    const int *i = func();
}

Answer 1

§12.2/5:

"A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full expression containing the call."

That means as i is being initialized, it's getting the address of a temporary object that does exist at that point. As soon as i is initialized, however, the temporary object will be destroyed, and i will become just another dangling pointer.

As such, yes, the function is valid -- but with the surrounding code as you've written it, any code you added afterward that attempted to dereference i would give undefined behavior.

Answer 2

Just because a pointer has a value doesn't mean it's a valid pointer.

In this case it holds an address which used to be that of x, and chances are that address still has the value 5, but it's not valid pointer and you can't count on that value being there.