英文:
Using std::enable_if with complex predicates
问题
使用 typename std::enable_if<IsVector<T>::value || IsList<T>::value, void>::type 作为约束条件时，它可以正常工作。如果使用 typename std::enable_if<IsVectorOrList<T>::value>::type，编译器会报错。
如何让它正常工作呢？
英文:
Having to use C++14 I'd like to use std::enable_if to make a function only being available given certain constraints. I do so using or_<> taken from the libc header type_traits instead of std::disjunction (no C++17 available in my environment) in order to move the constraint logic into a separate construct.
However, I seem to be missing something, because I can't manage to get it to compile.
My code:
#include <type_traits>#include <iostream>#include <vector>#include <list>template <typename T>struct IsVector : public std::false_type {};template <typename T>struct IsVector<std::vector<T>> : public std::true_type {};template <typename T>struct IsList : public std::false_type {};template <typename T>struct IsList<std::list<T>> : public std::true_type {};// taken from libc type_traitstemplate <bool, typename, typename>struct conditional;template <typename...>struct or_;template <>struct or_<> : public std::false_type {};template <typename _B1>struct or_<_B1> : public _B1 {};template <typename _B1, typename _B2>struct or_<_B1, _B2> : public conditional<_B1::value, _B1, _B2>::type {};template <typename _B1, typename _B2, typename _B3, typename... _Bn>struct or_<_B1, _B2, _B3, _Bn...> : public conditional<_B1::value, _B1, or_<_B2, _B3, _Bn...>>::type {};// ---template <typename T>struct IsVectorOrList : public or_<IsVector<T>, IsList<T>>::type {};template <typename T>typename std::enable_if<IsVector<T>::value || IsList<T>::value, void>::type// replacing with this line does not work//typename std::enable_if<IsVectorOrList<T>::value, void>::typefoo(const T& list){for (const auto& item : list){std::cout << item << std::endl;}}int main(){foo(std::vector<int>{17, 42});foo(std::list<float>{1.0, 2.71, 3.14});}
When using typename std::enable_if<IsVector<T>::value || IsList<T>::value, void>::type as constraint it works fine. If I use typename std::enable_if<IsVectorOrList<T>::value>::type the compiler complains:
traits.cpp:32:8: error: invalid use of incomplete type ‘struct conditional<true, IsVector<std::vector<int> >, IsList<std::vector<int> > >’32 | struct or_<_B1, _B2> : public conditional<_B1::value, _B1, _B2>::type {};| ^~~~~~~~~~~~~traits.cpp:20:8: note: declaration of ‘struct conditional<true, IsVector<std::vector<int> >, IsList<std::vector<int> > >’20 | struct conditional;| ^~~~~~~~~~~
How do I make it work?
答案1
得分: 2
以下是翻译好的部分：
"// taken from libc type_traits" -> "// 从 libc type_traits 中获取"
"The code declares conditional, but that class isn't defined anywhere in the posted snippet." -> "这段代码声明了 conditional，但该类在发布的片段中没有定义。"
"Removing those lines and using std::conditional the code works." -> "删除这些行并使用 std::conditional 代码可以正常工作。"
英文:
Here:
// taken from libc type_traitstemplate <bool, typename, typename>struct conditional;
The code declares conditional, but that class isn't defined anywhere in the posted snippet.
Removing those lines and using std::conditional the code works.
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