如何使用 std::enable_if_t 进行完美转发？

Question

I wrote a function receiving a variadic number of std::pairs that subtracts the 2nd element from the 1st element of each pair, and returns a tuple from the newly generated results like this:

template<typename... pairs, std::enable_if_t<((std::is_same_v<std::pair<int, int>, pairs>) && ...)>* = nullptr>
inline constexpr auto foo(pairs&& ...p) noexcept {
    return std::tuple((std::get<1>(std::forward<pairs>(p)) - std::get<0>(std::forward<pairs>(p)))  ...);
}

int main() {
    constexpr auto bar = foo(std::pair(4, 6), std::pair(3, 7), std::pair(1, 2));

    return 0;
}

This works perfectly fine. Unless I try to pass it a std::pair that isn't an rvalue, like this:

int main() {
    constexpr auto tup = std::pair(4, 6);
    constexpr auto bar = foo(tup, std::pair(3, 7), std::pair(1, 2));

    return 0;
}

If I try to do that, I get the following error:

no matching function for call to 'foo(const std::pair<int, int>&, std::pair<int, int>, std::pair<int, int>)'

How can I pass std::pairs both as lvalue and rvalue references to this function? I'm using C++17

Answer 1

When you pass Lvalue, pairs will keep reference in its type, so

std::is_same_v<std::pair<int, int>, pairs>

results in false.

You have to discard reference for example by using std::decay_t :

std::enable_if_t<((std::is_same_v<std::pair<int, int>, std::decay_t<pairs> >) && ...)>* = nullptr

Answer 2

When you pass an lvalue to the function, its corresponding type in the parameter pack is deduced as an lvalue reference due to how forwarding references work. You can use std::decay_t to remove all references and cv-qualifiers prior to your check:

std::enable_if_t<((std::is_same_v<std::pair<int, int>, std::decay_t<pairs>>) && ...)>