SFINAE only works for substituted template parameters, not any complication error (or it would be called CEINAE); therefore hash<T> for enums will fail to compile, given a non-enum, rather than being ignored. It's not possible to specialize hash<> for enums, since the primary template does not have en extra Enable template argument for use with enable_if. We therefore rename it to enum_hash<> and require users to explicitly define hash<MyEnum> as inheriting from it.
43 lines
1.3 KiB
C++
43 lines
1.3 KiB
C++
/*
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* This file is open source software, licensed to you under the terms
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* of the Apache License, Version 2.0 (the "License"). See the NOTICE file
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* distributed with this work for additional information regarding copyright
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* ownership. You may not use this file except in compliance with the License.
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*
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* You may obtain a copy of the License at
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*
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* http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing,
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* software distributed under the License is distributed on an
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* "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
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* KIND, either express or implied. See the License for the
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* specific language governing permissions and limitations
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* under the License.
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*/
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/*
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* Copyright (C) 2015 Cloudius Systems, Ltd.
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*/
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#pragma once
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/*
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* This header file defines a hash function for enum types, using the
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* standard hash function of the underlying type (such as int). This makes
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* it possible to inherit from this type to
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*/
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#include <type_traits>
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#include <functional>
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#include <cstddef>
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template <typename T>
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class enum_hash {
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static_assert(std::is_enum<T>::value, "must be an enum");
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public:
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std::size_t operator()(const T& e) const {
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using utype = typename std::underlying_type<T>::type;
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return std::hash<utype>()(static_cast<utype>(e));
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}
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};
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