一种模板专用化，可用于多个枚举值 [英] One template specialization for several enum values

问题描述

通常，如果我想通过枚举创建一个模板化(数据)类，我会写这样的东西

Normally if I want to have a templated (data) class by enum I would write something like this

enum class Modes : int
{
    m1 = 1,
    m2 = 2,
    m3 = 3
};

template <Modes M>
class DataHolder
{
};

template<>
class DataHolder<Modes::m1>
{
    public: int a = 4;
};

然后，如果我想对 Modes :: m1 使用与 Modes :: m2 相同的专业化，我会再次写同样的专业化.有没有一种方法可以为多个枚举值编写一个专门化的文字?我已经在SFINAE上尝试过了，但是我没有成功.

Then if I want the same specialization for the Modes::m1 as for the Modes::m2 I would write the same specialization again. Is there a way to write one specialization for several enum values? I have tried it with SFINAE, but I am not succesfull.

template <Modes M, typename = void>
class DataHolder
{
};

template<Modes M, typename = typename std::enable_if<M == Modes::m1 || M == Modes::m2>::type>
class DataHolder
{
    public: int a = 4;
};

这不能编译.特别是在我想对 Modes :: m3 进行不同的专业化之后.我已经尝试过在SO上找到许多类似的解决方案，但似乎没有任何解决方案.

This doesn't not compile. Especially, after I would like to carry on with different specialization for Modes::m3. I've tried many similiar solution found here on SO, but nothing seems to be solving the issue.

推荐答案

您应将 enable_if 放在显式的 DataHolder 专业名称中，该专业名称与默认名称相匹配.如果 enable_if 中的条件评估为 true ，则将选择专业化.

You should put the enable_if in an explicit specialization of DataHolder which matches the default one. The specialization will be chosen if the condition in the enable_if evaluates to true.

template <Modes M, typename = void>
class DataHolder
{
};

template<Modes M>
class DataHolder<M, typename std::enable_if<M == Modes::m1 || M == Modes::m2>::type>
{
    public: int a = 4;
};

int main()
{   
    DataHolder<Modes::m1> a; a.a;
    DataHolder<Modes::m3> b; /* b.a; */
}

godbolt.org上的实时示例

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