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For example, I have a class

class A

public:
 template<class T, class... Args>
 void set(Args&&... args);

private:
 std::shared_ptr<Member1Type> m_member1;
 std::shared_ptr<Member2Type> m_member2; // Member types are all different.
;

And I hope I can use it as

A a;
a.set<Member1Type>(args... to construct Member1Type);

which like

make_shared<T>(args...);

My question is how to link member type to the correct member in implementing the method. Thanks!

I'd populate a std::tuple<MemberType1*, MemberType2*, ...> in the ctor, so you can then use get<T*>(m_tuple) in A::set<T, Args...>

[edit]
Or as StoryTeller suggested, without extra members:

private:
 std::tuple <
 std::shared_ptr<Member1Type>,
 std::shared_ptr<Member2Type>
 > m_members;

You'd now need std::get<std::shared_ptr<T>>(m_members)

If you don't want to go the tuple approach, one way you could do this is to provide private getters for each member, overloaded on a tag type:

template <typename T> struct type_t ;
template <typename T> constexpr type_t<T> type;

class A

public:
 template<class T, class... Args>
 void set(Args&&... args) 
 get(type<T>) = std::make_shared<T>(std::forward<Args>(args)...);
 

private:
 auto& get(type_t<Member1Type>) return m_member1; 
 auto& get(type_t<Member2Type>) return m_member2; 

 std::shared_ptr<Member1Type> m_member1;
 std::shared_ptr<Member2Type> m_member2;
;

The auto& return avoids the need to write the type again, which already appears in the parameter list (kind of the same way you wouldn't repeat the type when writing a C++ cast).

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