将std::unique_ptr与lambda交换为deleter--GCC

swap std::unique_ptr with lambda as deleter -- GCC

本文关键字：lambda 交换 deleter--GCC ptr std unique      更新时间：2023-10-16

我们可以使用lambda作为std:：unique_ptr的deleter吗？事实上，我用叮当++做了这件事，它很高兴这样做。

我正在使用std::swap交换到std::unique_ptr<ObjType, decltyp(deleter)>;，其中auto deleter = [](struct addrinfo* ptr){if (ptr != nullptr) {freeaddrinfo(ptr);} };。Clang的swap似乎不需要拷贝分配运算符，但gcc的std:：swap需要，正如您在这些日志中看到的那样：

In file included from /usr/include/c++/4.8.1/memory:81:0,
                 from /home/zenol/proj/src/PROJ/TCPClient.cpp:28:
/usr/include/c++/4.8.1/bits/unique_ptr.h: In instantiation of ‘std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(std::unique_ptr<_Tp, _Dp>&&) [with _Tp = addrinfo; _Dp = Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0]’:
/usr/include/c++/4.8.1/bits/move.h:176:11:   required from ‘void std::swap(_Tp&, _Tp&) [with _Tp = std::unique_ptr<addrinfo, Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0>]’
/home/zenol/proj/src/Proj/SocketHelp.hpp:109:50:   required from ‘void Proj::retrieve_addresses(std::string, int, addrinfo&, addrinfo*&, T&, U) [with T = std::unique_ptr<addrinfo, Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0>; U = Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0; std::string = std::basic_string<char>]’
/home/zenol/proj/src/PROJ/TCPClient.cpp:65:49:   required from here
/usr/include/c++/4.8.1/bits/unique_ptr.h:193:16: erreur: use of deleted function ‘Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0& Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0::operator=(const Proj::TCPClient::connect(const Proj::SocketAddress&, int)::__lambda0&)’
  get_deleter() = std::forward<deleter_type>(__u.get_deleter());
                ^
/home/zenol/proj/src/Proj/TCPClient.cpp:56:21: note: a lambda closure type has a deleted copy assignment operator
     auto deleter = [](struct addrinfo* ptr)
                     ^

标准是什么？我能设法交换这两个std:：unique_ptr吗？它们是一种变通方法吗？（也许将lambda封装在std:：函数中？…）

编辑：这里有一个小例子，应该或多或少是一样的：

auto deleter = [](struct addrinfo* ptr)
{if (ptr != nullptr) {freeaddrinfo(ptr);} };
std::unique_ptr<struct addrinfo, decltype(deleter)>
resources_keeper(nullptr, deleter);
int main()
{
    decltype(resources_keeper) plouf1(nullptr, deleter);
    decltype(resources_keeper) plouf2(nullptr, deleter);
    std::swap(plouf1, plouf2);
    return 0;
}

错误：

In file included from /usr/include/c++/4.8.1/bits/stl_pair.h:59:0,
                 from /usr/include/c++/4.8.1/bits/stl_algobase.h:64,
                 from /usr/include/c++/4.8.1/memory:62,
                 from mini.cpp:1:
/usr/include/c++/4.8.1/bits/move.h: In instantiation of ‘void std::swap(_Tp&, _Tp&) [with _Tp = __lambda0]’:
/usr/include/c++/4.8.1/tuple:381:36:   required from ‘void std::_Tuple_impl<_Idx, _Head, _Tail ...>::_M_swap(std::_Tuple_impl<_Idx, _Head, _Tail ...>&) [with long unsigned int _Idx = 1ul; _Head = __lambda0; _Tail = {}]’
/usr/include/c++/4.8.1/tuple:382:35:   required from ‘void std::_Tuple_impl<_Idx, _Head, _Tail ...>::_M_swap(std::_Tuple_impl<_Idx, _Head, _Tail ...>&) [with long unsigned int _Idx = 0ul; _Head = addrinfo*; _Tail = {__lambda0}]’
/usr/include/c++/4.8.1/tuple:667:33:   required from ‘void std::tuple<_T1, _T2>::swap(std::tuple<_T1, _T2>&) [with _T1 = addrinfo*; _T2 = __lambda0]’
/usr/include/c++/4.8.1/tuple:1050:7:   required from ‘void std::swap(std::tuple<_Elements ...>&, std::tuple<_Elements ...>&) [with _Elements = {addrinfo*, __lambda0}]’
/usr/include/c++/4.8.1/bits/unique_ptr.h:269:21:   required from ‘void std::unique_ptr<_Tp, _Dp>::swap(std::unique_ptr<_Tp, _Dp>&) [with _Tp = addrinfo; _Dp = __lambda0]’
/usr/include/c++/4.8.1/bits/unique_ptr.h:484:7:   required from ‘void std::swap(std::unique_ptr<_Tp, _Dp>&, std::unique_ptr<_Tp, _Dp>&) [with _Tp = addrinfo; _Dp = __lambda0]’
mini.cpp:21:29:   required from here
/usr/include/c++/4.8.1/bits/move.h:176:11: erreur: use of deleted function ‘__lambda0& __lambda0::operator=(const __lambda0&)’
       __a = _GLIBCXX_MOVE(__b);
           ^
mini.cpp:9:17: note: a lambda closure type has a deleted copy assignment operator
 auto deleter = [](struct addrinfo* ptr)
                 ^
In file included from /usr/include/c++/4.8.1/bits/stl_pair.h:59:0,
                 from /usr/include/c++/4.8.1/bits/stl_algobase.h:64,
                 from /usr/include/c++/4.8.1/memory:62,
                 from mini.cpp:1:
/usr/include/c++/4.8.1/bits/move.h:177:11: erreur: use of deleted function ‘__lambda0& __lambda0::operator=(const __lambda0&)’
       __b = _GLIBCXX_MOVE(__tmp);
           ^

这与unique_ptr或tuple无关，您可以将错误减少为：

int main()
{
  auto deleter = []() { };
  auto del2 = deleter;
  deleter = static_cast<decltype(deleter)>(del2);
}

它用Clang编译，但用G++失败，给出了这个错误：

t.cc: In function ‘int main()’:
t.cc:5:11: error: use of deleted function ‘main()::<lambda()>& main()::<lambda()>::operator=(const main()::<lambda()>&)’
   deleter = static_cast<decltype(deleter)>(del2);
           ^
t.cc:3:19: note: a lambda closure type has a deleted copy assignment operator
   auto deleter = []() { };
                   ^

最后一个C++11标准在[expr.prim.lambda]/19:中说

与lambda表达式关联的闭包类型有一个已删除的（8.4.3）默认构造函数和一个已复制赋值运算符。它有一个隐式声明的复制构造函数（12.8），也可能有一个隐含声明的移动构造函数（12.8）

因此，类型是否可移动赋值取决于编译器。

扩展Jonathan Wakely的答案：

当您交换到unique_ptr时，您还必须交换它们的删除程序您看到的问题可以归结为：clang可以交换两个相同类型的lambda，gcc不能（正如Jonathan引用的那样，标准允许两者）。演示：

#include <utility>
int main() {
  auto f = [](){};
  auto g(f);
  std::swap(f, g);
}

这段代码可以使用clang，但无法使用gcc进行编译。（这没关系。）

这就是它发生的原因

我建议如下：

#include <memory>
#include <utility>
struct addrinfo { };
void freeaddrinfo(addrinfo* ) { }
struct deleter {
  void operator()(struct addrinfo* ptr) {
    if (ptr != nullptr)
      freeaddrinfo(ptr);
  }
};
using resources_keeper = std::unique_ptr<struct addrinfo, deleter>;
int main() {
    resources_keeper plouf1(nullptr);
    resources_keeper plouf2(nullptr);
    std::swap(plouf1, plouf2);
    return 0;
}

请注意，代码变得更干净、可读性更强。

如果你绝对必须用lambdas来解决这个问题，那么也许你可以尝试类似的技巧性：只交换指针，而不交换deleter。

#include <iostream>
#include <memory>
#include <utility>
using namespace std;
template <class T, class D>
void swap_pointers_but_not_deleters(unique_ptr<T,D>& x, unique_ptr<T,D>& y) noexcept {
  T* x_ptr = x.release();
  x.reset(y.release());
  y.reset(x_ptr);
}
int main() {
  auto deleter = [](int* p){ delete p; };
  unique_ptr<int,decltype(deleter)> a(new int(1),deleter);
  unique_ptr<int,decltype(deleter)> b(new int(2),deleter);
  swap_pointers_but_not_deleters(a, b);
  cout << "a = " << *a << ", b = " << *b << endl;
}

虽然这段代码看起来有效，但我真的不喜欢它。我建议第一个不使用lambdas的解决方案。

我可以用以下代码重现类似的错误：

struct A
{
    A() = default;
    A(A&&) = default;
    //A & operator=(A&&) = default;
    A(A const & ) = delete;
};
int main()
{
    A a, b;
    std::swap(a,b);
}

取消对移动赋值运算符的注释，错误就会消失。我猜gcc不允许对lambda进行移动赋值（我使用的是4.7.2版本）。将lambda更改为实际的函数或函子，你应该会没事的。

事实证明，您可以用lambda来解决它，只要它们可以转换为函数指针（lambda什么都不捕获）。

#include <memory>
#include <utility>
struct addrinfo { };
void freeaddrinfo(addrinfo* ) { }
auto deleter = [](struct addrinfo* ptr) {
  if (ptr != nullptr)
    freeaddrinfo(ptr);
};
using resources_keeper = std::unique_ptr<struct addrinfo, void(*)(struct addrinfo*)>;
int main() {
    resources_keeper plouf1(nullptr,deleter);
    resources_keeper plouf2(nullptr,deleter);
    std::swap(plouf1, plouf2);
    return 0;
}

然而，我仍然更喜欢我的另一个具有结构的解决方案。它可能是最有效的一个（由于内联），其次是这里提供的解决方案。如果deleter实现真的很简单，那么传递一个重量级的std::function对我来说似乎有些过头了。这些性能考虑因素是否重要，由探查器来判断。

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