用lambda捕获完全转发的变量

template<typename T> void doSomething(T&& mStuff)
{
    auto lambda([&mStuff]{ doStuff(std::forward<T>(mStuff)); });
    lambda();
}

Is it correct to capture the perfectly-forwarded mStuff variable with the &mStuff syntax?

使用&mStuff语法捕获完全转发的mStuff变量是否正确?

Or is there a specific capture syntax for perfectly-forwarded variables?

或者对于完全转发的变量是否有特定的捕获语法?

EDIT: What if the perfectly-forwarded variable is a parameter pack?

编辑:如果完全转发的变量是一个参数包呢?

3 个解决方案

#1

Is it correct to capture the perfectly-forwarded mStuff variable with the &mStuff syntax?

使用&mStuff语法捕获完全转发的mStuff变量是否正确?

Yes, assuming that you don't use this lambda outside doSomething. Your code captures mStuff per reference and will correctly forward it inside the lambda.

是的，假设你不在doSomething外使用这个lambda。您的代码捕获每个引用的mStuff并将它正确地转发到lambda中。

For mStuff being a parameter pack it suffices to use a simple-capture with a pack-expansion:

mStuff作为参数包，只需使用简单的-capture和package -expansion:

template <typename... T> void doSomething(T&&... mStuff)
{
    auto lambda = [&mStuff...]{ doStuff(std::forward<T>(mStuff)...); };
}

The lambda captures every element of mStuff per reference. The closure-object saves an lvalue reference for to each argument, regardless of its value category. Perfect forwarding still works; In fact, there isn't even a difference because named rvalue references would be lvalues anyway.

lambda捕获每个引用的mStuff的每个元素。closure-object为每个参数保存一个lvalue引用，而不考虑其值类别。完美转发仍然工作;事实上，甚至没有区别，因为命名rvalue引用无论如何都是lvalue。

#2

Yes you can do perfect capturing, but not directly. You will need to wrap the type in another class:

是的，你可以做完美的捕捉，但不是直接的。您需要将类型包装在另一个类中:

#define REQUIRES(...) class=std::enable_if_t<(__VA_ARGS__)>

template<class T>
struct wrapper
{
    T value;
    template<class X, REQUIRES(std::is_convertible<T, X>())>
    wrapper(X&& x) : value(std::forward<X>(x))
    {}

    T get() const
    {
        return std::move(value);
    }
};

template<class T>
auto make_wrapper(T&& x)
{
    return wrapper<T>(std::forward<T>(x));
}

Then pass them as parameters to a lambda that returns a nested lambda that captures the parameters by value:

然后将它们作为参数传递给返回嵌套lambda，该lambda按值捕获参数:

template<class... Ts>
auto do_something(Ts&&... xs)
{
    auto lambda = [](auto... ws)
    {
        return [=]()
        {
            // Use `.get()` to unwrap the value
            some_other_function(ws.get()...);
        };
    }(make_wrapper(std::forward<Ts>(xs)...));

    lambda();
}

#3

To make the lambda valid outside the scope where it's created, you need a wrapper class that handles lvalues and rvalues differently, i.e., keeps a reference to an lvalue, but makes a copy of (by moving) an rvalue.

要使lambda在创建范围之外有效，您需要一个包装器类来处理不同的lvalue和rvalue，例如。，保留对lvalue的引用，但复制(通过移动)一个rvalue。

Header file capture.h:

头文件capture.h:

#pragma once

#include <type_traits>
#include <utility>

template < typename T >
class capture_wrapper
{
   static_assert(not std::is_rvalue_reference<T>{},"");
   std::remove_const_t<T> mutable val_;
public:
   constexpr explicit capture_wrapper(T&& v)
      noexcept(std::is_nothrow_move_constructible<std::remove_const_t<T>>{})
   :val_(std::move(v)){}
   constexpr T&& get() const noexcept { return std::move(val_); }
};

template < typename T >
class capture_wrapper<T&>
{
   T& ref_;
public:
   constexpr explicit capture_wrapper(T& r) noexcept : ref_(r){}
   constexpr T& get() const noexcept { return ref_; }
};

template < typename T >
constexpr typename std::enable_if<
   std::is_lvalue_reference<T>{},
   capture_wrapper<T>
>::type
capture(std::remove_reference_t<T>& t) noexcept
{
   return capture_wrapper<T>(t);
}

template < typename T >
constexpr typename std::enable_if<
   std::is_rvalue_reference<T&&>{},
   capture_wrapper<std::remove_reference_t<T>>
>::type
capture(std::remove_reference_t<T>&& t)
   noexcept(std::is_nothrow_constructible<capture_wrapper<std::remove_reference_t<T>>,T&&>{})
{
   return capture_wrapper<std::remove_reference_t<T>>(std::move(t));
}

template < typename T >
constexpr typename std::enable_if<
   std::is_rvalue_reference<T&&>{},
   capture_wrapper<std::remove_reference_t<T>>
>::type
capture(std::remove_reference_t<T>& t)
   noexcept(std::is_nothrow_constructible<capture_wrapper<std::remove_reference_t<T>>,T&&>{})
{
   return capture_wrapper<std::remove_reference_t<T>>(std::move(t));
}

Example/test code that shows it works. Note that the "bar" example shows how one can use std::tuple<...> to work around the lack of pack expansion in lambda capture initializer, useful for variadic capture.

示例/测试代码，显示其工作。注意，“bar”示例显示了如何使用std::tuple<…>在lambda捕获初始化器中解决包展开不足的问题，这对于可变捕获非常有用。

#include <cassert>
#include <tuple>
#include "capture.h"

template < typename T >
auto foo(T&& t)
{
   return [t = capture<T>(t)]()->decltype(auto)
   {
      auto&& x = t.get();
      return std::forward<decltype(x)>(x);
      // or simply, return t.get();
   };
}

template < std::size_t... I, typename... T >
auto bar_impl(std::index_sequence<I...>, T&&... t)
{
   static_assert(std::is_same<std::index_sequence<I...>,std::index_sequence_for<T...>>{},"");
   return [t = std::make_tuple(capture<T>(t)...)]()
   {
      return std::forward_as_tuple(std::get<I>(t).get()...);
   };
}
template < typename... T >
auto bar(T&&... t)
{
   return bar_impl(std::index_sequence_for<T...>{}, std::forward<T>(t)...);
}

int main()
{
   static_assert(std::is_same<decltype(foo(0)()),int&&>{}, "");
   assert(foo(0)() == 0);

   auto i = 0;
   static_assert(std::is_same<decltype(foo(i)()),int&>{}, "");
   assert(&foo(i)() == &i);

   const auto j = 0;
   static_assert(std::is_same<decltype(foo(j)()),const int&>{}, "");
   assert(&foo(j)() == &j);

   const auto&& k = 0;
   static_assert(std::is_same<decltype(foo(std::move(k))()),const int&&>{}, "");
   assert(foo(std::move(k))() == k);

   auto t = bar(0,i,j,std::move(k))();
   static_assert(std::is_same<decltype(t),std::tuple<int&&,int&,const int&,const int&&>>{}, "");
   assert(std::get<0>(t) == 0);
   assert(&std::get<1>(t) == &i);
   assert(&std::get<2>(t) == &j);
   assert(std::get<3>(t) == k and &std::get<3>(t) != &k);

}