如何扩展可变参数模板 class

Question

Simplified I'm trying the following:

template<typename T>
struct SomeObject
{
    T value;
};

template<typename... Ts>
struct Container : public SomeObject<Ts>...
{
    void init()
    {
        (initObject<Ts>())...; // not working: how to execute initObject for each type
    }

    template<typename T>
    void initObject()
    {
        SomeObject<T>::value = new T();
        // do some other stuff
    }
};

class A {};
class B {};

int main()
{
    auto c = new Container<A, B>();
    c->init();
}

In my use case I cannot do the initialization in the constructor. So how can I get the init method to properly expand for all types?

Answer 1

In c++17, you can do it with a fold expression over the comma operator

(initObject<Ts>(), ...);

before c++17 you could do something like

std::initializer_list<int>{(initObject<Ts>(), 0)...};

Note the comma experssion in the initializer list. It is there because initObject returns void , and you cannot have an initializer_list of void s

Here is another more general solution. It is too complex for this case, but may be useful.

template<typename... Ts>
struct Container : public SomeObject<Ts>...
{
    void init()
    {
        initObjects<Ts...>(); // not working: how to execute initObject for each type
    }

    template <typename ...Args>
    struct object_initializer;

    template <typename First, typename ...Rest>
    struct object_initializer<First, Rest...> {
        void operator ()() {
            SomeObject<First> obj{First()};
            object_initializer<Rest...>()();
        }
    };

    template <typename First>
    struct object_initializer<First> {
        void operator ()() {
            SomeObject<First> obj{First()};
        }
    };

    template<typename ...Args>
    void initObjects() {
        object_initializer<Args...>()();
    }
};