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如何在std::apply中封装std::make_unique?

unique-ptr tuples c++

bers · 技术社区 · 1 年前

以下内容未编译:

#include <memory>
#include <tuple>
#include <utility>

int main()
{
    std::apply(std::make_unique<int>, std::make_tuple(1));
    
    return 0;
}

错误消息:

In file included from /usr/include/c++/11/bits/unique_ptr.h:37,
                 from /usr/include/c++/11/memory:76,
                 from main.cpp:1:
/usr/include/c++/11/tuple: In instantiation of âconstexpr decltype(auto) std::__apply_impl(_Fn&&, _Tuple&&, std::index_sequence<_Idx ...>) [with _Fn = std::unique_ptr<int, std::default_delete<int> > (&)(); _Tuple = std::tuple<int>; long unsigned int ..._Idx = {0}; std::index_sequence<_Idx ...> = std::integer_sequence<long unsigned int, 0>]â:
/usr/include/c++/11/tuple:1869:31:   required from âconstexpr decltype(auto) std::apply(_Fn&&, _Tuple&&) [with _Fn = std::unique_ptr<int, std::default_delete<int> > (&)(); _Tuple = std::tuple<int>]â
main.cpp:7:15:   required from here
/usr/include/c++/11/tuple:1858:27: error: no matching function for call to â__invoke(std::unique_ptr > (&)(), std::__tuple_element_t<0, std::tuple >)â
 1858 |       return std::__invoke(std::forward<_Fn>(__f),
      |              ~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
 1859 |                            std::get<_Idx>(std::forward<_Tuple>(__t))...);
      |                            ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In file included from /usr/include/c++/11/tuple:41,
                 from /usr/include/c++/11/bits/unique_ptr.h:37,
                 from /usr/include/c++/11/memory:76,
                 from main.cpp:1:
/usr/include/c++/11/bits/invoke.h:90:5: note: candidate: âtemplate constexpr typename std::__invoke_result<_Functor, _ArgTypes>::type std::__invoke(_Callable&&, _Args&& ...)â
   90 |     __invoke(_Callable&& __fn, _Args&&... __args)
      |     ^~~~~~~~
/usr/include/c++/11/bits/invoke.h:90:5: note:   template argument deduction/substitution failed:
/usr/include/c++/11/bits/invoke.h: In substitution of âtemplate<class _Callable, class ... _Args> constexpr typename std::__invoke_result<_Functor, _ArgTypes>::type std::__invoke(_Callable&&, _Args&& ...) [with _Callable = std::unique_ptr<int, std::default_delete<int> > (&)(); _Args = {int}]â:
/usr/include/c++/11/tuple:1858:27:   required from âconstexpr decltype(auto) std::__apply_impl(_Fn&&, _Tuple&&, std::index_sequence<_Idx ...>) [with _Fn = std::unique_ptr<int, std::default_delete<int> > (&)(); _Tuple = std::tuple<int>; long unsigned int ..._Idx = {0}; std::index_sequence<_Idx ...> = std::integer_sequence<long unsigned int, 0>]â
/usr/include/c++/11/tuple:1869:31:   required from âconstexpr decltype(auto) std::apply(_Fn&&, _Tuple&&) [with _Fn = std::unique_ptr<int, std::default_delete<int> > (&)(); _Tuple = std::tuple<int>]â
main.cpp:7:15:   required from here
/usr/include/c++/11/bits/invoke.h:90:5: error: no type named âtypeâ in âstruct std::__invoke_result > (&)(), int>â

这怎么能解决?

1 回复 | 直到 1 年前

NathanOliver 1 年前

C++17及更早版本

std::make_unique<int> 具有类型 std::unique_ptr<int>() 这是一个返回 std::unique_ptr<int> 并且不采用任何参数。这不是您想要的,因为您正在向函数传递值。

你需要的是 std::make_unique<int, int> 具有类型 std::unique_ptr<int>(int) 这就是你想要的。第一个模板参数用于的类型 unique_ptr 并且每个附加类型用于make函数的参数类型。

C++20+

自C++20以来 make_unique 不是可寻址函数,因此不再允许在可移植代码中使用此技术。相反,您需要利用lambda来实现间接级别,如

std::apply([](auto&& var){ return std::make_unique<int>(decltype(var)(var)); }, std::make_tuple(1));