In really simple terms, std::move returns an argument as an rvalue reference while std::forward returns either an lvalue reference or an rvalue reference based on how the argument was passed in to the current function. While this is fairly obvious once you get the hang of it, it can be quite tricky to grasp earlier on. The easiest way to see how they work differently is to try some code such as the following:
void Show(int&&)
{
cout << "int&& called" << endl;
}
void Show(int&)
{
cout << "int& called" << endl;
}
template<typename T> void Foo(T&& x)
{
cout << "straight: ";
Show(x);
cout << "move: ";
Show(move(x));
cout << "forward: ";
Show(forward<T>(x));
cout << endl;
}
int _tmain(void)
{
Foo(10);
int x=10;
Foo(x);
return 0;
}
The output of the above code is:
straight: int& called move: int&& called forward: int&& called straight: int& called move: int&& called forward: int& called
When Foo is called with an rvalue, T&& is deduced as int&&. But the variable itself is an lvalue within that function. So the straight call calls the Show(int&) overload – no surprises there. Both the move-call and the forward-call calls Show(int&&) as expected. In the 2nd case, Foo is called with an lvalue, and again the straight call will go to Show(int&) while the move-call will go to Show(int&&). Now here’s where forward comes in handy, it will go to Show(int&) because that’s what T has been deduced to in this instance of the function call (perfect forwarding). If you are wondering why, the collapsing rule for T&& collapses it to T& during template argument type deduction.
Really helpful post…………..