[英]How is std::shared_ptr polymorphic?
I'm developing some container class, which has similar idea: holding pointer inside. 我正在开发一些容器类,它具有类似的想法:在内部保存指针。
#include <iostream>
template<class T>
class Container
{
public:
Container ( )
{
pointer = new T ( );
}
~Container ( )
{
delete pointer;
}
T* operator->( )
{
return pointer;
}
private:
T* pointer;
};
struct Base
{
virtual void who ( )
{
std::cout << "Base" << std::endl;
}
};
struct Child : public Base
{
virtual void who ( ) override
{
std::cout << "Child" << std::endl;
}
};
void testContainer ( Container<Base> c )
{
c->who ( );
}
void testSharedPtr ( std::shared_ptr<Base> s )
{
s->who ( );
}
int main ( )
{
Container<Child> child;
std::shared_ptr<Child> sharedChild;
testSharedPtr ( sharedChild );
testContainer ( child );
}
This code fails to compile: error C2664: 'void test(Container<Base> &)' : cannot convert argument 1 from 'Container<Child>' to 'Container<Base> &'
此代码无法编译: error C2664: 'void test(Container<Base> &)' : cannot convert argument 1 from 'Container<Child>' to 'Container<Base> &'
However with std::shared_ptr<Base>
, instead of Container
, everything works fine. 但是使用std::shared_ptr<Base>
而不是Container
,一切正常。 So question is: 所以问题是:
Is std::shared_ptr
-like polymorphysm implementable? 像std::shared_ptr
的 polymorphysm是否可以实现? Or is this feature, which was somehow hardcoded in C++? 还是此功能(在C ++中以某种方式进行了硬编码)? Sorry for my primitive language. 对不起,我的原始语言。
Implement the following constructor: 实现以下构造函数:
template<class Other,
class = typename std::enable_if<std::is_convertible<Other*, T*>::value>::type>
Container(const Container<Other>& _Other)
{
pointer = _Other.pointer;
}
This constructor will only be enabled if Other*
is convertible to T*
, which is obviosly the case if T
is the base class of Other
. 仅当Other*
可转换为T*
时才启用此构造函数,如果T
是Other
的基类,则情况明显。
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