如何轻松地将 C++ 枚举映射到字符串
我在使用的一些库头文件中有一堆枚举类型，我想要一种将枚举值转换为用户字符串的方法，反之亦然.
I have a bunch of enum types in some library header files that I'm using, and I want to have a way of converting enum values to user strings - and vice-versa.
RTTI 不会为我做这件事，因为用户字符串"需要比枚举更具可读性.
RTTI won't do it for me, because the 'user strings' need to be a bit more readable than the enumerations.
一个蛮力的解决方案是一堆这样的函数，但我觉得这有点太像 C 了.
A brute force solution would be a bunch of functions like this, but I feel that's a bit too C-like.
enum MyEnum {VAL1, VAL2,VAL3};
String getStringFromEnum(MyEnum e)
{
switch e
{
case VAL1: return "Value 1";
case VAL2: return "Value 2";
case VAL1: return "Value 3";
default: throw Exception("Bad MyEnum");
}
}
我有一种使用模板的优雅解决方案的直觉，但我还不能完全理解它.
I have a gut feeling that there's an elegant solution using templates, but I can't quite get my head round it yet.
更新:感谢您的建议 - 我应该明确指出枚举是在第三方库标头中定义的，因此我不想更改它们的定义.
UPDATE: Thanks for suggestions - I should have made clear that the enums are defined in a third-party library header, so I don't want to have to change the definition of them.
我现在的直觉是避免使用模板并执行以下操作:
My gut feeling now is to avoid templates and do something like this:
char * MyGetValue(int v, char *tmp); // implementation is trivial
#define ENUM_MAP(type, strings) char * getStringValue(const type &T)
{
return MyGetValue((int)T, strings);
}
; enum eee {AA,BB,CC}; - exists in library header file
; enum fff {DD,GG,HH};
ENUM_MAP(eee,"AA|BB|CC")
ENUM_MAP(fff,"DD|GG|HH")
// To use...
eee e;
fff f;
std::cout<< getStringValue(e);
std::cout<< getStringValue(f);
推荐答案
如果您希望枚举名称本身为字符串，请参阅 这篇文章.否则， std::map 会很好地工作.(将字符串文字复制到映射中的 std::strings 没有意义)
If you want the enum names themselves as strings, see this post.
Otherwise, a std::map<MyEnum, char const*> will work nicely. (No point in copying your string literals to std::strings in the map)
对于额外的语法糖，这里是如何编写 map_init 类.目标是允许
For extra syntactic sugar, here's how to write a map_init class. The goal is to allow
std::map<MyEnum, const char*> MyMap;
map_init(MyMap)
(eValue1, "A")
(eValue2, "B")
(eValue3, "C")
;
函数template 返回一个 map_init_helper.map_init_helper<T> 存储一个 T&，并定义了微不足道的 map_init_helper&operator()(typename T::key_type const&, typename T::value_type const&).(从 operator() 返回 *this 允许链接 operator()，就像 operator<<< onstd::ostreams)
The function template <typename T> map_init(T&) returns a map_init_helper<T>.
map_init_helper<T> stores a T&, and defines the trivial map_init_helper& operator()(typename T::key_type const&, typename T::value_type const&). (Returning *this from operator() allows the chaining of operator(), like operator<< on std::ostreams)
template<typename T> struct map_init_helper
{
T& data;
map_init_helper(T& d) : data(d) {}
map_init_helper& operator() (typename T::key_type const& key, typename T::mapped_type const& value)
{
data[key] = value;
return *this;
}
};
template<typename T> map_init_helper<T> map_init(T& item)
{
return map_init_helper<T>(item);
}
由于函数和辅助类是模板化的，您可以将它们用于任何地图或类似地图的结构.IE.它还可以将条目添加到 std::unordered_map
Since the function and helper class are templated, you can use them for any map, or map-like structure. I.e. it can also add entries to std::unordered_map
如果您不喜欢编写这些帮助程序，boost::assign 提供了开箱即用的相同功能.
If you don't like writing these helpers, boost::assign offers the same functionality out of the box.
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