Implicit conversion of shared_ptr - c++

Question

I have two shared_ptrs of the classes U and T where T is the base of U.
It is no problem to do an implicit conversion from shared_ptr<U> to shared_ptr<T>.
But is is also possible to do a conversion from shared_ptr<T> to shared_ptr<U> ?
I tried the sugested solution:
class T {
public:
virtual ~T() {}
protected:
void fillData() = 0;
};
class Timpl : public T
{
public:
virtual ~Timpl() {}
protected:
virtual void fillData() = 0;
};
class U : public Timpl {
public:
virtual ~U() {}
protected:
virtual void fillData() {...} // fill
};
typedef shared_ptr<T> TPtr
typedef shared_ptr<U> UPtr
TPtr tp = std::make_shared<U>();
UPtr up = std::static_pointer_cast<U>(tp); //<-- error here :
error: no matching function for call to 'static_pointer_cast(TPtr)'
note: template std::__shared_ptr<_Tp1, _Lp> std::static_pointer_cast(const std::__shared_ptr<_Tp2, _Lp>&)
note: template argument deduction/substitution failed:
note: 'TPtr {aka boost::shared_ptr}' is not derived from 'const std::__shared_ptr<_Tp2, _Lp>'
Solution for this issue:
mixing of std::shared_ptr with boost::shared_ptr is no good idea.

Answer 1

Yes, use static_pointer_cast:
#include <memory>
struct T { virtual ~T() {} };
struct U : T {};
std::shared_ptr<T> pt = std::make_shared<U>(); // *pt is really a "U"
auto pu = std::static_pointer_cast<U>(pt);
There's also a matching std::dynamic_pointer_cast which returns a null pointer if the conversion isn't possible.