The std::forward function as the std::move function aims at implementing move semantics in C++. The function takes a forwarding reference. According to the T template parameter, std::forward identifies whether an lvalue or an rvalue reference has been passed to it and returns a corresponding kind of reference. std::forward helps to implement perfect forwarding. This mechanism implies that objects passed to the function as lvalue expressions should be copied, and objects passed to the function as rvalue expressions should be moved.
If you assign an rvalue reference to some ref variable, then ref is a named entity. Therefore, its category is lvalue, despite the fact that ref is an rvalue reference. Therefore, for ref copy semantics is used instead of move semantics.
The std::forward function solves this problem. Let's consider the following example:
template <typename T>
void foo(T &&arg)
{
std::vector<int> var = arg;
....
}
std::vector<int> vect(1'000'000, 1);
foo(std::move(vect));
That's what happens in the code fragment:
Now let's use std::forward:
template <typename T>
void foo(T &&arg)
{
std::vector<int> var = std::forward<T>(arg);
....
}
std::vector<int> vect(1'000'000, 1);
foo(std::move(vect));
Here's what happens:
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