c ++ lambdas如何从上层范围捕获可变参数包

Question

我研究了通用的 lambdas，并稍微修改了示例，所以我的 lambda 应该捕获上层 lambda 的可变参数包。因此，基本上，作为 (auto&&...) 提供给上层 lambda 的内容应该以某种方式在 [=] 块中捕获。

（完美转发又是一个问题，我很好奇这里有没有可能？）

#include <iostream>
#include<type_traits>
#include<utility>


// base case
void doPrint(std::ostream& out) {}

template <typename T, typename... Args>
void doPrint(std::ostream& out, T && t, Args && ... args)
{
    out << t << " ";                // add comma here, see below
    doPrint(out, std::forward<Args&&>(args)...);
}

int main()
{
    // generic lambda, operator() is a template with one parameter
    auto vglambda = [](auto printer) {
        return [=](auto&&... ts) // generic lambda, ts is a parameter pack
        {
            printer(std::forward<decltype(ts)>(ts)...);
            return [=] {  // HOW TO capture the variadic ts to be accessible HERE ↓
                printer(std::forward<decltype(ts)>(ts)...); // ERROR: no matchin function call to forward
            }; // nullary lambda (takes no parameters)
        };
    };
    auto p = vglambda([](auto&&...vars) {
        doPrint(std::cout, std::forward<decltype(vars)>(vars)...);
    });
    auto q = p(1, 'a', 3.14,5); // outputs 1a3.14

    //q(); //use the returned lambda "printer"

}

Answer 1

The perfect forwarding is another question, I'm curious is it possible here at all?

嗯...在我看来完美转发是的问题

ts... 的捕获效果很好，如果您在内部 lambda 中进行更改，

printer(std::forward<decltype(ts)>(ts)...);

和

printer(ts...);

程序编译。

问题是通过值（使用 [=]）捕获 ts...，它们变成 const 值和 printer()（这是一个接收 auto&&...vars) 接收引用（& 或 &&）的 lambda。

你可以在下面的函数中看到同样的问题

void bar (int &&)
 { }

void foo (int const & i)
 { bar(std::forward<decltype(i)>(i)); }

从 clang++ 我得到

tmp_003-14,gcc,clang.cpp:21:4: error: no matching function for call to 'bar'
 { bar(std::forward<decltype(i)>(i)); }
   ^~~
tmp_003-14,gcc,clang.cpp:17:6: note: candidate function not viable: 1st argument
      ('const int') would lose const qualifier
void bar (int &&)
     ^

另一种解决问题的方法是捕获 ts... 作为引用（因此 [&]）而不是作为值。

Answer 2

在C++20中完美捕获

template <typename ... Args>
auto f(Args&& ... args){
    return [... args = std::forward<Args>(args)]{
        // use args
    };
}

C++17 和 C++14 解决方法

在 C++17 中，我们可以使用元组的变通方法：

template <typename ... Args>
auto f(Args&& ... args){
    return [args = std::make_tuple(std::forward<Args>(args) ...)]()mutable{
        return std::apply([](auto&& ... args){
            // use args
        }, std::move(args));
    };
}

不幸的是 std::apply 是 C++17，在 C++14 中你可以自己实现它或者用 boost::hana 做类似的事情：

namespace hana = boost::hana;

template <typename ... Args>
auto f(Args&& ... args){
    return [args = hana::make_tuple(std::forward<Args>(args) ...)]()mutable{
        return hana::unpack(std::move(args), [](auto&& ... args){
            // use args
        });
    };
}

通过函数 capture_call:

来简化解决方法可能会很有用

#include <tuple>

// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
    return [
        lambda = std::forward<Lambda>(lambda),
        capture_args = std::make_tuple(std::forward<Args>(args) ...)
    ](auto&& ... original_args)mutable{
        return std::apply([&lambda](auto&& ... args){
            lambda(std::forward<decltype(args)>(args) ...);
        }, std::tuple_cat(
            std::forward_as_tuple(original_args ...),
            std::apply([](auto&& ... args){
                return std::forward_as_tuple< Args ... >(
                    std::move(args) ...);
            }, std::move(capture_args))
        ));
    };
}

这样使用：

#include <iostream>

// returns a callable object without parameters
template <typename ... Args>
auto f1(Args&& ... args){
    return capture_call([](auto&& ... args){
        // args are perfect captured here
        // print captured args via C++17 fold expression
        (std::cout << ... << args) << '\n';
    }, std::forward<Args>(args) ...);
}

// returns a callable object with two int parameters
template <typename ... Args>
auto f2(Args&& ... args){
    return capture_call([](int param1, int param2, auto&& ... args){
        // args are perfect captured here
        std::cout << param1 << param2;
        (std::cout << ... << args) << '\n';
    }, std::forward<Args>(args) ...);
}

int main(){
    f1(1, 2, 3)();     // Call lambda without arguments
    f2(3, 4, 5)(1, 2); // Call lambda with 2 int arguments
}

这里是 capture_call 的 C++14 实现：

#include <tuple>

// Implementation detail of a simplified std::apply from C++17
template < typename F, typename Tuple, std::size_t ... I >
constexpr decltype(auto)
apply_impl(F&& f, Tuple&& t, std::index_sequence< I ... >){
    return static_cast< F&& >(f)(std::get< I >(static_cast< Tuple&& >(t)) ...);
}

// Implementation of a simplified std::apply from C++17
template < typename F, typename Tuple >
constexpr decltype(auto) apply(F&& f, Tuple&& t){
    return apply_impl(
        static_cast< F&& >(f), static_cast< Tuple&& >(t),
        std::make_index_sequence< std::tuple_size<
            std::remove_reference_t< Tuple > >::value >{});
}

// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
    return [
        lambda = std::forward<Lambda>(lambda),
        capture_args = std::make_tuple(std::forward<Args>(args) ...)
    ](auto&& ... original_args)mutable{
        return ::apply([&lambda](auto&& ... args){
            lambda(std::forward<decltype(args)>(args) ...);
        }, std::tuple_cat(
            std::forward_as_tuple(original_args ...),
            ::apply([](auto&& ... args){
                return std::forward_as_tuple< Args ... >(
                    std::move(args) ...);
            }, std::move(capture_args))
        ));
    };
}

capture_call 按值捕获变量。完美意味着尽可能使用移动构造函数。这是一个 C++17 代码示例，以便更好地理解：

#include <tuple>
#include <iostream>
#include <boost/type_index.hpp>


// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
    return [
        lambda = std::forward<Lambda>(lambda),
        capture_args = std::make_tuple(std::forward<Args>(args) ...)
    ](auto&& ... original_args)mutable{
        return std::apply([&lambda](auto&& ... args){
            lambda(std::forward<decltype(args)>(args) ...);
        }, std::tuple_cat(
            std::forward_as_tuple(original_args ...),
            std::apply([](auto&& ... args){
                return std::forward_as_tuple< Args ... >(
                    std::move(args) ...);
            }, std::move(capture_args))
        ));
    };
}

struct A{
    A(){
        std::cout << "  A::A()\n";
    }

    A(A const&){
        std::cout << "  A::A(A const&)\n";
    }

    A(A&&){
        std::cout << "  A::A(A&&)\n";
    }

    ~A(){
        std::cout << "  A::~A()\n";
    }
};

int main(){
    using boost::typeindex::type_id_with_cvr;

    A a;
    std::cout << "create object end\n\n";

    [b = a]{
        std::cout << "  type of the capture value: "
          << type_id_with_cvr<decltype(b)>().pretty_name()
          << "\n";
    }();
    std::cout << "value capture end\n\n";

    [&b = a]{
        std::cout << "  type of the capture value: "
          << type_id_with_cvr<decltype(b)>().pretty_name()
          << "\n";
    }();
    std::cout << "reference capture end\n\n";

    [b = std::move(a)]{
        std::cout << "  type of the capture value: "
          << type_id_with_cvr<decltype(b)>().pretty_name()
          << "\n";
    }();
    std::cout << "perfect capture end\n\n";

    [b = std::move(a)]()mutable{
        std::cout << "  type of the capture value: "
          << type_id_with_cvr<decltype(b)>().pretty_name()
          << "\n";
    }();
    std::cout << "perfect capture mutable lambda end\n\n";

    capture_call([](auto&& b){
        std::cout << "  type of the capture value: "
          << type_id_with_cvr<decltype(b)>().pretty_name()
          << "\n";
    }, std::move(a))();
    std::cout << "capture_call perfect capture end\n\n";
}

输出：

  A::A()
create object end

  A::A(A const&)
  type of the capture value: A const
  A::~A()
value capture end

  type of the capture value: A&
reference capture end

  A::A(A&&)
  type of the capture value: A const
  A::~A()
perfect capture end

  A::A(A&&)
  type of the capture value: A
  A::~A()
perfect capture mutable lambda end

  A::A(A&&)
  type of the capture value: A&&
  A::~A()
capture_call perfect capture end

  A::~A()

捕获值的类型在capture_call版本中包含&&，因为我们必须通过引用访问内部元组中的值，而支持捕获的语言支持直接访问该值.

Answer 3

您可以使用 std::bind 将可变参数绑定到您的 lambda（对于 pre-C++20解）：

template <typename... Args>
auto f(Args&&... args){

    auto functional = [](auto&&... args) { /* lambda body */ };
    return std::bind(std::move(functional), std::forward<Args>(args)...);
}

c ++ lambdas如何从上层范围捕获可变参数包

c++ lambdas how to capture variadic parameter pack from the upper scope

c++

lambda

c++14

variadic-templates

generic-lambda