shared_ptr initialization

Question

A member is defined as ``` std::shared_ptr > exit_to; ``` which points to additional data shared among others. When try to initiate the pointer "exit_to". The correct way is ``` node_knot.exit_to = std::make_shared >(); ``` But it's in another file and I'd like to keep the pointer type consistent, something like this: ``` node_knot.exit_to = std::make_shared(); ``` But won't compile: ``` /usr/include/c++/4.6/bits/shared_ptr_base.h:798:54: error: '__p' declared as a pointer to a reference of type 'std::array, 6> &' __shared_ptr(const __shared_ptr<_Tp1, _Lp>& __r, _Tp* __p) ^ /usr/include/c++/4.6/bits/shared_ptr.h:93:31: note: in instantiation of template class 'std::__shared_ptr, 6> &, 1>' requested here class shared_ptr : public __shared_ptr<_Tp> ^ ../node_booker.h:757:20: note: in instantiation of template class 'std::shared_ptr

Accepted Answer

You need to remove the reference from the type you are passing to `make_shared`. The following should work: ``` node_knot.exit_to = std::make_shared::type>(); ```

shared_ptr initialization

Solution

Problem