c++ - 如何使用自定义删除器复制 unique_ptr

Question

如果我使用自定义删除器制作 unique_ptr 的拷贝，我会收到编译错误。请有人帮助我。

#include <iostream>
#include<memory>
#include <algorithm>
using namespace std;

auto del = [](int *p) { cout <<"obj deleted "<<endl;delete p;};
int main()
{
   unique_ptr<int, decltype(del)> p1(new int(10), del);
   unique_ptr<int,decltype(del)> p2;
   p2 = std::move(p1);
}

错误:

C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\tuple||In instantiation of 'constexpr std::_Head_base<_Idx, _Head, true>::_Head_base() [with unsigned int _Idx = 1u; _Head = <lambda(int*)>]':|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\tuple|353|required from 'constexpr std::_Tuple_impl<_Idx, _Head>::_Tuple_impl() [with unsigned int _Idx = 1u; _Head = <lambda(int*)>]'|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\tuple|202|required from 'constexpr std::_Tuple_impl<_Idx, _Head, _Tail ...>::_Tuple_impl() [with unsigned int _Idx = 0u; _Head = int*; _Tail = {<lambda(int*)>}]'|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\tuple|602|required from 'constexpr std::tuple<_T1, _T2>::tuple() [with _T1 = int*; _T2 = <lambda(int*)>]'|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\bits\unique_ptr.h|158|required from 'constexpr std::unique_ptr<_Tp, _Dp>::unique_ptr() [with _Tp = int; _Dp = <lambda(int*)>]'|
F:\3d\C++CodeProject\Hello\main.cpp|10|required from here|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\tuple|59|error: use of deleted function '<lambda(int*)>::<lambda>()'|
F:\3d\C++CodeProject\Hello\main.cpp|6|note: a lambda closure type has a deleted default constructor|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\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 = int; _Dp = <lambda(int*)>]':|
F:\3d\C++CodeProject\Hello\main.cpp|11|required from here|
C:\Program Files (x86)\CodeBlocks\MinGW\lib\gcc\mingw32\5.1.0\include\c++\bits\unique_ptr.h|252|error: use of deleted function '<lambda(int*)>&<lambda(int*)>::operator=(const<lambda(int*)>&)'|
F:\3d\C++CodeProject\Hello\main.cpp|6|note: a lambda closure type has a deleted copy assignment operator|
||=== Build failed: 2 error(s), 8 warning(s) (0 minute(s), 1 second(s)) ===|

Answer 1

p2 声明的问题是 lambda 的类型不是 default constructible .因此，指针类型的每个声明都需要将删除器的实例传递给它。

您可以通过将 deleter lambda 替换为默认可构造的显式仿函数结构来解决此问题。然后，指针类型的每个实例都将能够创建自己的删除器对象实例。

struct MyDeleter
{
    void operator()(int* p){cout << "deleted" << endl; delete p;}
};

int main()
{
    //don't need to specify an instance of the deleter since it is default_constructible.
    unique_ptr<int, MyDeleter> p1(new int(10)); 
    unique_ptr<int, MyDeleter> p2;
    p2 = std::move(p1);
}

编辑:正如@super 所说，分配行的问题还在于(pre-c++20)lambdas 也不是 CopyAssignable .我在上面发布的仿函数方法解决了这两个问题。