is any difference between std::forward and std::forward?

Question

are these functions equivalent?

template 
void foo(T && t)
{
    bar(std::forward(t));
}

template 
void foo2(         


        

Answer 1

The accepted answer does not solve the problem in title completely.

A macro argument preserves the type of the expression. A forwarding parameter in a template does not. This means t in foo2 (as a forwarding function parameter) has the type T&& (because this is the forwarding template parameter), but it can be something different when the macro is in other contexts. For example:

using T = int;
T a = 42;
T&& t(std::move(a));
foo(MY_FORWARD(t)); // Which foo is instantiated?

Note here t is not an xvalue, but an lvalue. With std::forward(t), which is equivalent to std::forward(t), t would be forwarded as an lvalue. However, with MY_FORWARD(t), which is equivalent to std::forward(t), t would be forwarded as an xvalue. This contextual-dependent difference is sometime desired when you have to deal with some declared variables with rvalue reference types (not forwarding paramter even they may look like similar in syntax).