std::forward of a function passed via universal reference?
c++, c++11, lambda, perfect-forwarding, universal-reference
Solution
There is a difference if `Function`'s `operator()` has ref qualifiers. With `std::forward`, the value category of the argument is propagated, without it, the value category is lost, and the function will always be called as an l-value. Live Example.
#include <iostream>
struct Fun {
void operator()() & {
std::cout << "L-Value\n";
}
void operator()() && {
std::cout << "R-Value\n";
}
};
template <class Function>
void apply(Function&& function) {
function();
}
template <class Function>
void apply_forward(Function&& function) {
std::forward<Function>(function)();
}
int main () {
apply(Fun{}); // Prints "L-Value\n"
apply_forward(Fun{}); // Prints "R-Value\n"
}
Problem
Consider the two following: ``` template <class Function> void apply(Function&& function) { std::forward<Function>(function)(); } ``` and ``` template <class Function> void apply(Function&& function) { function(); } ``` In what case is there a difference, and what concrete difference is it ?