CODEWITHSUNDEEP
c++c++11templatesc++14
user1899020
user1899020

Reputation: 13575

How to make set<T>(args...) method?

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!

Upvotes: 2

Views: 127

Answers (3)

user10794579
user10794579

Reputation: 1

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Upvotes: -1

MSalters
MSalters

Reputation: 179981

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)

Upvotes: 7

Barry
Barry

Reputation: 303347

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).

Upvotes: 3

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