Initialize std::shared_ptr<T> from std::shared_ptr<void>

Question

I run into it when trying to "copy" from std::shared_ptr<void> to templated std::shared_ptr<T>. It looks like this:

template < >
class TDynamic< void >
{
protected:
    std::shared_ptr< void > m_pointer;
public:
    // Here i got constructors, functions etc
    template < typename U >
    TDynamic< U > SwitchType()
    {
        TDynamic< U > returnValue;
        returnValue.m_pointer = std::shared_ptr< U >(m_pointer); // error here
    }
};

The error says this:

c:\...path_here...\tdynamic.hpp(426): error C2440: 'type cast': cannot convert from 'const std::shared_ptr< void >' to 'std::shared_ptr< T >'
1>          with
1>          [
1>              T=grim::Actor
1>          ]
1>  c:\...path_here...\tdynamic.hpp(426): note: No constructor could take the source type, or constructor overload resolution was ambiguous

Compiler & IDE: Visual Studio 2015 Community

This is necessary to me for this situation. Lets say I have class "Actor" and derived class "APlayer"

class Actor
{
public:
};
class APlayer
    : public Actor
{
public:
};

int main()
{ // some code here
    TDynamic< APlayer > test1(new APlayer);
    TDynamic< void > test2(test1);
    TDynamic< Actor > test3(test2); // error here
}

I could paste the whole code here but its like 550 lines and most of it does not matter. I just need to copy that std::shared_ptr.

Answer 1

You can use static_pointer_cast:

returnValue.m_pointer = static_pointer_cast<U>(m_pointer);

It's up to you to guarantee that this cast is legal (it's defined to be as legal as static_cast<U*>(m_pointer.get())).