使用从模板继承的声明的构造函数的正确语法 [duplicate]

问：

我在命名空间中有一个模板化基类（不确定这是否相关，因此请包含它以防万一）。我正在尝试使用构造函数的声明扩展类，并调用以下方法之一：using

#include <iostream>
#include <memory>

namespace ns
{

template <typename T>
class Base
{
public:
    Base(const std::shared_ptr<const T> x): p{std::move(x)} {}
    Base(const T &x): Base{std::shared_ptr<const T>(&x)} {}
    Base(T &&x): Base{std::make_shared<const T>(std::move(x))} {}

    T getP() { return *p; }
private:
    std::shared_ptr<T> p;
};

} // ns

template <typename T>
class Test : public ns::Base<T>
{
public:
    using ns::Base::Base;

    void print()
    {
        std::cout << getP() << std::endl;
    }
};

int main()
{
    Test<int> t{3};
    t.print();
}

鉴于，我收到以下错误：using ns::Base::Base;

snip.cpp:26:15: error: ‘template<class T> class ns::Base’ used without template arguments
   26 |     using ns::Base::Base;
      |               ^~~~

将其更改为通过构造函数行，但找不到方法的定义：using ns::Base<T>::Base;

snip.cpp: In member function ‘void Test<T>::print()’:
snip.cpp:30:22: error: there are no arguments to ‘getP’ that depend on a template parameter, so a declaration of ‘getP’ must be available [-fpermissive]
   30 |         std::cout << getP() << std::endl;
      |                      ^~~~
snip.cpp:30:22: note: (if you use ‘-fpermissive’, G++ will accept your code, but allowing the use of an undeclared name is deprecated)
snip.cpp: In instantiation of ‘ns::Base<T>::Base(std::shared_ptr<const _Tp>) [with T = int]’:
snip.cpp:13:62:   required from ‘ns::Base<T>::Base(T&&) [with T = int]’
snip.cpp:26:24:   required from here
snip.cpp:11:59: error: no matching function for call to ‘std::shared_ptr<int>::shared_ptr(<brace-enclosed initializer list>)’
   11 |     Base(const std::shared_ptr<const T> x): p{std::move(x)} {}
      |                                                           ^

如何在扩展中使用模板化基类，以便可以重用基构造函数并调用方法？