如何轉發可變的lambda

這是我嘗試編譯的代碼的淡化示例：

#include <iostream>
#include <functional>

template <class F>
auto foo(F&& fun)
{
    return [callback = std::forward<F>(fun)](auto&&... args) {
        std::invoke(callback, std::forward<decltype(args)>(args)...);
    };
}

int main()
{
    std::string cur("running"), target("ok");
    
    foo([s1 = cur, s2 = target](std::string const& arg) /*mutable*/ {
        if (s1 == arg)
        {
            std::cout << s1 << std::endl;
        }
    })("not ok");
    
    return 0;
}

簡單地說，我有一個foo接受可呼叫物件的函式，并且應該從它們構建一個新的可呼叫物件。就示例而言，上面我只是呼叫fun引數，但在實際情況下，對可呼叫物件進行了一些修飾，并將結果放入在某些條件下呼叫此類“操作”的資料結構中。

這個例子編譯并運行得很好。這個問題時，試圖通過可變lambda運算式來體現foo。當我取消注釋mutable上面的關鍵字時，出現以下編譯錯誤：

main.cpp: In instantiation of 'foo<main()::<lambda(const string&)> >(main()::<lambda(const string&)>&&)::<lambda(auto:1&& ...)> [with auto:1 = {const char (&)[7]}]':

main.cpp:21:7:   required from here

main.cpp:8:20: error: no matching function for call to 'invoke(const main()::<lambda(const string&)>&, const char [7])'

    8 |         std::invoke(callback, std::forward<decltype(args)>(args)...);

      |         ~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In file included from main.cpp:2:

/usr/local/include/c  /11.2.0/functional:94:5: note: candidate: 'template<class _Callable, class ... _Args> std::invoke_result_t<_Callable, _Args ...> std::invoke(_Callable&&, _Args&& ...)'

   94 |     invoke(_Callable&& __fn, _Args&&... __args)

      |     ^~~~~~

/usr/local/include/c  /11.2.0/functional:94:5: note:   template argument deduction/substitution failed:

In file included from /usr/local/include/c  /11.2.0/bits/move.h:57,

                 from /usr/local/include/c  /11.2.0/bits/nested_exception.h:40,

                 from /usr/local/include/c  /11.2.0/exception:148,

                 from /usr/local/include/c  /11.2.0/ios:39,

                 from /usr/local/include/c  /11.2.0/ostream:38,

                 from /usr/local/include/c  /11.2.0/iostream:39,

                 from main.cpp:1:

/usr/local/include/c  /11.2.0/type_traits: In substitution of 'template<class _Fn, class ... _Args> using invoke_result_t = typename std::invoke_result::type [with _Fn = const main()::<lambda(const string&)>&; _Args = {const char (&)[7]}]':

/usr/local/include/c  /11.2.0/functional:94:5:   required by substitution of 'template<class _Callable, class ... _Args> std::invoke_result_t<_Callable, _Args ...> std::invoke(_Callable&&, _Args&& ...) [with _Callable = const main()::<lambda(const string&)>&; _Args = {const char (&)[7]}]'

main.cpp:8:20:   required from 'foo<main()::<lambda(const string&)> >(main()::<lambda(const string&)>&&)::<lambda(auto:1&& ...)> [with auto:1 = {const char (&)[7]}]'

main.cpp:21:7:   required from here

/usr/local/include/c  /11.2.0/type_traits:2933:11: error: no type named 'type' in 'struct std::invoke_result<const main()::<lambda(const string&)>&, const char (&)[7]>'

 2933 |     using invoke_result_t = typename invoke_result<_Fn, _Args...>::type;

      |           ^~~~~~~~~~~~~~~

知道這是為什么嗎？我也可以foo接受可變的 lambda 運算式嗎？

uj5u.com熱心網友回復：

只需添加mutable到 lambda 中foo：

template <class F>
auto foo(F&& fun)
{
    return [callback = std::forward<F>(fun)](auto&&... args) mutable {
                                                             //^^^
        std::invoke(callback, std::forward<decltype(args)>(args)...);
    };
}

標籤：C 拉姆达 模板元编程

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