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c++ - 将 void * 转换为类型的 shared_ptr

问题描述

在下面的代码中，我试图将一个 void* 转换为一个类型的 shared_ptr：

#include <iostream>
#include <memory>
class A
{
public:
A(){l=0;}
int l;
void Show() { std::cout << l << "\n";}
};

void PrintA(void *aptr)
{
std::shared_ptr<A> a1;
a1.reset(aptr);
a1->Show();
}

int main()
{
std::shared_ptr<A> a(new A());
PrintA(a.get());
}

但我得到以下编译错误：

$ c++ -std=c++14 try20.cpp
In file included from C:/tools/mingw64/x86_64-w64-mingw32/include/c++/bits/shared_ptr.h:52:0,
                 from C:/tools/mingw64/x86_64-w64-mingw32/include/c++/memory:82,
                 from try20.cpp:2:
C:/tools/mingw64/x86_64-w64-mingw32/include/c++/bits/shared_ptr_base.h: In instantiation of 'std::__shared_ptr<_Tp, _Lp>::__shared_ptr(_Tp1*) [with _Tp1 = void; _Tp = A; __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]':
C:/tools/mingw64/x86_64-w64-mingw32/include/c++/bits/shared_ptr_base.h:1023:4:   required from 'void std::__shared_ptr<_Tp, _Lp>::reset(_Tp1*) [with _Tp1 = void; _Tp = A; __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]'
try20.cpp:14:14:   required from here
C:/tools/mingw64/x86_64-w64-mingw32/include/c++/bits/shared_ptr_base.h:871:39: error: invalid conversion from 'void*' to 'A*' [-fpermissive]
         : _M_ptr(__p), _M_refcount(__p)
                                       ^
C:/tools/mingw64/x86_64-w64-mingw32/include/c++/bits/shared_ptr_base.h:874:4: error: static assertion failed: incomplete type
    static_assert( !is_void<_Tp1>::value, "incomplete type" );

如何将 void 指针转换为类型的共享指针？

标签： c++c++11

解决方案

假设您不能更改 的声明PrintA，您的PrintA定义应如下所示：

void PrintA(void *aptr)
{
    A* a1 = reinterpret_cast<A*>(aptr);
    a1->Show();
}

当你传递给它一个指针。由于您不收回所有权，因此无需创建std::shared_ptr. 如果您需要共享所有权，则必须修改A以允许使用share_from_this.

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