完美地用于通用 lambda 中的自动 &&

Question

下一个泛型lambda用于保护线程之间共享资源的任何操作:

auto mutexed = [mtx(std::mutex{})](auto &&fn, auto &&...args) mutable {
std::unique_lock lo(mtx);
return fn(args...);
//return std::forward<decltype(fn)>(fn)(std::forward<decltype(args)>(args)...);
};

decltype(fn)对完全转发无效。它适用于值和右值引用。但不适合左值的refs。
写decltype((fn))可以用左值引用，但不能用右值引用。

所以这个问题。

如何使其完美地转发引用和右值引用?

让线程不破坏stdout的例子:https://gcc.godbolt.org/z/KsWM6Pq6x

#include <iostream>
#include <thread>
#include <chrono>
#include <mutex>
using namespace std::chrono_literals;
using std::cout, std::clog, std::cerr, std::endl;
int main() {
cout << "Hello, World!" << endl;
auto mutexed = [mtx(std::mutex{})](auto &&fn, auto &&...args) mutable {
std::unique_lock lo(mtx);
//return fn(args...);
return std::forward<decltype(fn)>(fn)(std::forward<decltype(args)>(args)...);
};

int counter = 0;
auto t1 = std::thread( [&](){ 
while(1){ 
std::this_thread::sleep_for(10ms);
// mutexed( [](int&&cnt){ cout << cnt++ << endl; }, std::move(counter) ); //FAILS HERE
mutexed( [](int&cnt){ cout << cnt++ << endl; }, counter );  //lval ref works fine
}
});
auto t2 = std::thread([&]{ 
while(1){
std::this_thread::sleep_for(10ms);
mutexed( []{ cout << "done long_operation_2" << endl;});}
});
t1.join();
t2.join();
}

---

c++ 11,14,17和c++ 20,23的工作解

int main() {
auto lambda20 = []<class F, class...Ts>(F &&fn, Ts &&...args) {
return std::forward<F>(fn)(std::forward<Ts>(args)...);
};
auto lambda14 = [](auto &&fn, auto &&...args) {
return std::forward<
std::conditional_t<
std::is_rvalue_reference_v<decltype(fn)>,
typename std::remove_reference_t<decltype(fn)>,
decltype(fn)>
>(fn)(
std::forward<
std::conditional_t<std::is_rvalue_reference<decltype(args)>::value,
typename std::remove_reference<decltype(args)>::type,
decltype(args)
>>(args)...);
};
int inter = 20;
lambda20([](int x) { cout << "asdf20   x" << endl; }, inter);
lambda20([](int &x) { cout << "asdf20  &x" << endl; }, inter);
lambda20([](int &&x) { cout << "asdf20 &&x" << endl; }, std::move(inter));
lambda14([](int x) { cout << "asdf14   x" << endl; }, inter);
lambda14([](int &x) { cout << "asdf14  &x" << endl; }, inter);
lambda14([](int &&x) { cout << "asdf14 &&x" << endl; }, std::move(inter));
return 0;
}

c++ pre-20解决方案8岁by Scott Meyers https://scottmeyers.blogspot.com/2013/05/c14-lambdas-and-perfect-forwarding.html

Answer 1

c++ 20及以后版本

auto lambda20 = []<class F, class...Ts>(F &&fn, Ts &&...args) {
return std::forward<F>(fn)(std::forward<Ts>(args)...);
};

c++ pre20: c++ 11,14,17

auto lambda14 = [](auto &&fn, auto &&...args) {
return std::forward<
std::conditional_t<
std::is_rvalue_reference_v<decltype(fn)>,
typename std::remove_reference_t<decltype(fn)>,
decltype(fn)>
>(fn)(
std::forward<
std::conditional_t<std::is_rvalue_reference<decltype(args)>::value,
typename std::remove_reference<decltype(args)>::type,
decltype(args)
>>(args)...);
};

例子

#include <iostream>
using namespace std;
int main() {
auto lambda20 = []<class F, class...Ts>(F &&fn, Ts &&...args) {
return std::forward<F>(fn)(std::forward<Ts>(args)...);
};
auto lambda14 = [](auto &&fn, auto &&...args) {
return std::forward<
std::conditional_t<
std::is_rvalue_reference_v<decltype(fn)>,
typename std::remove_reference_t<decltype(fn)>,
decltype(fn)>
>(fn)(
std::forward<
std::conditional_t<std::is_rvalue_reference<decltype(args)>::value,
typename std::remove_reference<decltype(args)>::type,
decltype(args)
>>(args)...);
};
int inter = 20;
lambda20([](int x) { cout << "asdf20   x" << endl; }, inter);
lambda20([](int &x) { cout << "asdf20  &x" << endl; }, inter);
lambda20([](int &&x) { cout << "asdf20 &&x" << endl; }, std::move(inter));
lambda14([](int x) { cout << "asdf14   x" << endl; }, inter);
lambda14([](int &x) { cout << "asdf14  &x" << endl; }, inter);
lambda14([](int &&x) { cout << "asdf14 &&x" << endl; }, std::move(inter));
return 0;
}

c++ pre-20解决方案8岁by Scott Meyers https://scottmeyers.blogspot.com/2013/05/c14-lambdas-and-perfect-forwarding.html