C++ 不能使用枚举类作为 unordered_map 键
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Can't use enum class as unordered_map key
提问by Appleshell
I have a class containing an enum class.
我有一个包含枚举类的类。
class Shader {
public:
enum class Type {
Vertex = GL_VERTEX_SHADER,
Geometry = GL_GEOMETRY_SHADER,
Fragment = GL_FRAGMENT_SHADER
};
//...
Then, when I implement the following code in another class...
然后，当我在另一个类中实现以下代码时......
std::unordered_map<Shader::Type, Shader> shaders;
...I get a compile error.
...我收到一个编译错误。
...usr/lib/c++/v1/type_traits:770:38:
Implicit instantiation of undefined template 'std::__1::hash<Shader::Type>'
What is causing the error here?
是什么导致了这里的错误？
回答by Daniel
I use a functor object to calculate hash of enum class:
我使用一个函子对象来计算散列enum class：
struct EnumClassHash
{
template <typename T>
std::size_t operator()(T t) const
{
return static_cast<std::size_t>(t);
}
};
Now you can use it as 3rd template-parameter of std::unordered_map:
现在您可以将其用作 的第三个模板参数std::unordered_map：
enum class MyEnum {};
std::unordered_map<MyEnum, int, EnumClassHash> myMap;
So you don't need to provide a specialization of std::hash, the template argument deduction does the job. Furthermore, you can use the word usingand make your own unordered_mapthat use std::hashor EnumClassHashdepending on the Keytype:
所以你不需要提供 的特化std::hash，模板参数推导就可以完成这项工作。此外，您可以使用这个词using并根据类型unordered_map使用std::hash或制作自己的词：EnumClassHashKey
template <typename Key>
using HashType = typename std::conditional<std::is_enum<Key>::value, EnumClassHash, std::hash<Key>>::type;
template <typename Key, typename T>
using MyUnorderedMap = std::unordered_map<Key, T, HashType<Key>>;
Now you can use MyUnorderedMapwith enum classor another type:
现在您可以使用MyUnorderedMapwithenum class或其他类型：
MyUnorderedMap<int, int> myMap2;
MyUnorderedMap<MyEnum, int> myMap3;
Theoretically, HashTypecould use std::underlying_typeand then the EnumClassHashwill not be necessary. That could be something like this, but I haven't tried yet:
理论上，HashType可以使用std::underlying_type然后EnumClassHash就没有必要了。这可能是这样的，但我还没有尝试过：
template <typename Key>
using HashType = typename std::conditional<std::is_enum<Key>::value, std::hash<std::underlying_type<Key>::type>, std::hash<Key>>::type;
If using std::underlying_typeworks, could be a very good proposal for the standard.
如果使用std::underlying_type作品，可能是标准的一个很好的建议。
回答by David Stone
This was considered a defect in the standard, and was fixed in C++14: http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#2148
这被认为是标准中的一个缺陷，并在 C++14 中得到了修复：http: //www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#2148
This is fixed in the version of libstdc++ shipping with gcc as of 6.1: https://gcc.gnu.org/bugzilla/show_bug.cgi?id=60970.
从 6.1 开始，这在带有 gcc 的 libstdc++ 版本中已修复：https://gcc.gnu.org/bugzilla/show_bug.cgi?id =60970。
It was fixed in clang's libc++ in 2013: http://lists.cs.uiuc.edu/pipermail/cfe-commits/Week-of-Mon-20130902/087778.html
它于 2013 年在 clang 的 libc++ 中得到修复：http: //lists.cs.uiuc.edu/pipermail/cfe-commits/Week-of-Mon-20130902/087778.html
回答by denim
A very simple solution would be to provide a hash function object like this:
一个非常简单的解决方案是提供一个像这样的哈希函数对象：
std::unordered_map<Shader::Type, Shader, std::hash<int> > shaders;
That's all for an enum key, no need to provide a specialization of std::hash.
这就是枚举键的全部内容，无需提供 std::hash 的特化。
回答by CS Pei
When you use std::unordered_map, you know you need a hash function. For built-in or STLtypes, there are defaults available, but not for user-defined ones. If you just need a map, why don't you try std::map?
当您使用 时std::unordered_map，您知道您需要一个哈希函数。对于内置或STL类型，有可用的默认值，但不适用于用户定义的。如果你只需要一张地图，为什么不试试std::map呢？
回答by Daniel Frey
As KerrekSB pointed out, you need to provide a specialization of std::hashif you want to use std::unordered_map, something like:
正如 KerrekSB 指出的那样，std::hash如果要使用std::unordered_map，则需要提供专业化，例如：
namespace std
{
template<>
struct hash< ::Shader::Type >
{
typedef ::Shader::Type argument_type;
typedef std::underlying_type< argument_type >::type underlying_type;
typedef std::hash< underlying_type >::result_type result_type;
result_type operator()( const argument_type& arg ) const
{
std::hash< underlying_type > hasher;
return hasher( static_cast< underlying_type >( arg ) );
}
};
}
回答by user3080602
Add this to header defining MyEnumClass:
将此添加到定义 MyEnumClass 的标头中：
namespace std {
template <> struct hash<MyEnumClass> {
size_t operator() (const MyEnumClass &t) const { return size_t(t); }
};
}
回答by Vladimir Shutow
Try
尝试
std::unordered_map<Shader::Type, Shader, std::hash<std::underlying_type<Shader::Type>::type>> shaders;
