In perfect forwarding, std::forward is used to convert the named rvalue references t1 and t2 to unnamed rvalue references. What is the purpose of doing that? How would that affect the called function inner if we leave t1 & t2 as lvalues?
template <typename T1, typename T2>
void outer(T1&& t1, T2&& t2)
{
inner(std::forward<T1>(t1), std::forward<T2>(t2));
}
From another viewpoint, when dealing with rvalues in a universal reference assignment, it may be desirable to preserve the type of a variable as it is. For example
auto&& x = 2; // x is int&&
auto&& y = x; // But y is int&
auto&& z = std::forward<decltype(x)>(x); // z is int&&
Using std::forward, we ensured z exactly has the same type as x.
Moreover, std::forward doesn't affect lvalue references:
int i;
auto&& x = i; // x is int&
auto&& y = x; // y is int&
auto&& z = std::forward<decltype(x)>(x); // z is int&
Still z has the same type as x.
So, back to your case, if the inner function has two overloads for int& and int&&, you want to pass variables like z assignment not y one.
The types in the example can be assessed via:
std::cout<<is_same_v<int&,decltype(z)>;
std::cout<<is_same_v<int&&,decltype(z)>;