模板参数扣除与强类型枚举 [英] template argument deduction with strongly-typed enumerations

问题描述

如果我有一个正常（弱）枚举，我可以使用其枚举值作为非类型模板参数，如下：

If I have a normal (weak) enumeration, I can use its enumerated values as non-type template parameters, like so:

enum { Cat, Dog, Horse };

template <int Val, typename T> bool magic(T &t)
{
    return magical_traits<Val>::invoke(t);
}

并调用： magic< Cat> t）

我可以看到，如果我有一个强类型枚举并且不想硬编码枚举类型，我最终以：

as far as I can see, if I have a strongly-typed enumeration and don't want to hard-code the enumeration type, I end up with:

enum class Animal { Cat, Dog, Horse };

template <typename EnumClass, EnumClass EnumVal, typename T> bool magic(T &t)
{
    return magical_traits<EnumVal>::invoke(t);
}

现在我必须写： magic& ，Animal :: Cat>（t），这似乎是多余的。

and now I have to write: magic<Animal, Animal::Cat>(t), which seems redundant.

有任何方法可以避免输出枚举类和值，短于

Is there any way to avoid typing out both the enum class and the value, short of

#define MAGIC(E, T) (magic<decltype(E), E>(T));

推荐答案

很抱歉，

取出宏，放入一个可怕的命名标题，同事的清理脚本。希望最好的。

Take the macro, put it into a scary named header and protect it from your colleague's cleanup script. Hope for the best.

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