Question

我有一套模版的构件,其中有一些缩略语。是否有做以下工作的途径?

template< typename A >
struct someClass{
    enum class State{
        sA,
        sB,
        sC
    }

    static const std::array<State,4> the_states = {{
        State::sA,
        State::sB,
        State::sC
    }};

};

Answer 1

No. 只有静态综合数据成员才能在某一类别内初步形成。

然而,你可以这样做......

template< typename A >
struct someClass
{
    enum State
    {
        sA,
        sB,
        sC
    };

    static const std::array<const State,4> the_states;
};

template<typename A>
const std::array<const someClass::State,4> someClass<A>::the_states = 
{
    someClass::State::sA, 
    someClass::State::sB, 
    someClass::State::sC 
};

Answer 2

#include <iostream>
#include <array>

template< typename A >
struct someClass{
    enum class State {
        sA,
        sB,
        sC
    };

    static const std::array<State,3> the_states;    
};

template<typename A>
const std::array<typename someClass<A>::State,3> someClass<A>::the_states = {
    someClass<A>::State::sA,
    someClass<A>::State::sB,
    someClass<A>::State::sC
};

int main() {
    for( auto i : someClass<int>::the_states) {
        switch(i) {
            case someClass<int>::State::sA:
                std::cout << "sA" << std::endl;
                break;
            case someClass<int>::State::sB:
                std::cout << "sB" << std::endl;
                break;
            case someClass<int>::State::sC:
                std::cout << "sC" << std::endl;
                break;
        }
    }
}

请注意,你可以以零敲碎打的方式终止名单,因为0人不能被转换成一个大类国家。

巴哈教徒,当我对真实的Dave先生的回答置之不理的时候。

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