交換std :: unique_ptr lambda as deleter - GCC

Question

我們可以使用lambda作爲刪除與std :: unique_ptr嗎？實際上，我是用clang ++來做的，很高興能這樣做。

我正在使用std::swap轉換爲std::unique_ptr<ObjType, decltyp(deleter)>;，其中auto deleter = [](struct addrinfo* ptr){if (ptr != nullptr) {freeaddrinfo(ptr);} };。鐺的交換似乎並不需要一個拷貝賦值運算符，但gcc的化std :: swap一樣，你可以在這些日誌中看到：

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) 
        ^

什麼標準說？我可以設法消除這兩個std :: unique_ptr？他們是一種解決方法嗎？ （也許封裝拉姆達一個std ::函數裏面......？）

編輯： 這裏是一個小例子，應該是或多或少同樣的事情：

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; 
} 

錯誤：

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); 
     ^

Answer 1

要擴大Jonathan Wakely's answer：

當你換到unique_ptr S，你也有交換他們的刪除器。你看到的問題歸結爲：鐺可以交換兩個相同類型的lambda，gcc不能（並且該標準允許Jonathan引用它）。示範：

#include <utility> 

int main() { 
    auto f = [](){}; 
    auto g(f); 
    std::swap(f, g); 
} 

此代碼與clang一起使用，但無法使用gcc進行編譯。 （這是行）

這就是爲什麼它發生。

---

我建議如下：

請注意，代碼變得更清潔，更具有可讀性爲好。

---

如果絕對有解決這個與lambda表達式，那麼也許你可以嘗試一些的hackish這樣的：只交換指針而不是刪除者。

#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; 
} 

雖然這段代碼似乎工作，我真的不喜歡它。我建議第一個不使用lambdas的解決方案。

Answer 2

我可以重現一個類似的錯誤用下面的代碼：

struct A 
{ 
    A() = default; 
    A(A&&) = default; 
    //A & operator=(A&&) = default; 
    A(A const &) = delete; 
}; 

int main() 
{ 
    A a, b; 
    std::swap(a,b); 
} 

取消對移動賦值操作符的註釋並且錯誤消失。我猜gcc不允許移動分配的lambas（我正在使用版本4.7.2）。將lambda改爲一個實際的函數或仿函數，你應該沒問題。

Answer 3

這有什麼好做unique_ptr或tuple，可以將誤差減少到這一點：

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

與鏘編譯但無法與G ++，給這個錯誤：

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 = []() { }; 
       ^

的最後的C++ 11標準在[expr.prim.lambda]/19中說：

The closure type associated with a lambda-expression has a deleted (8.4.3) default constructor and a deleted copy assignment operator. It has an implicitly-declared copy constructor (12.8) and may have an implicitly-declared move constructor (12.8).

所以它是由編譯器決定的，類型是可移動賦值還是不賦值。

Answer 4

事實證明，只要它們可以轉換爲函數指針（lambdas不捕獲任何東西），就可以用lambdas來解決它。

#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; 
} 

不過，我還是喜歡我的other solution與結構更好。這可能是效率最高的一種（感謝內聯），接下來是此處介紹的解決方案。如果刪除器實現非常簡單，那麼傳遞一個重量級的std::function看起來對我來說太過分了。無論這些性能考慮事項是否重要，都是要分析人員的工作。