Question

我正在使用一些模板技巧来推断某些函数的返回类型，但是由于某种原因，这段代码无法编译。我希望您的帮助能够理解错误的原因，并在可能的情况下如何解决。

可以找到变体类here。

我有以下代码：

g++  -std=c++11 -Ideps -I.   -c -o Value/Value.o Value/Value.cpp
In file included from Value/Value.cpp:1:0:
Value/Value.hpp: In instantiation of ‘Value::ResolveReturnConvertType<To>             Value::convertTo() const [with To = int; Value::ResolveReturnConvertType<To> =     int; typename ConverterAdaptor<To>::ConverterType = NumericConverter<int>]’:
Value/Value.cpp:27:33:   required from here
Value/Value.hpp:88:3: error: no matching function for call to   ‘mapbox::util::variant<std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > >, unsigned int, float, bool,     std::shared_ptr<const std::vector<std::shared_ptr<const Value>,     std::allocator<std::shared_ptr<const Value> > > >, std::shared_ptr<const     std::unordered_map<std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > >, std::shared_ptr<const Value>,     std::hash<std::shared_ptr<const std::basic_string<char, std::char_traits<char>,     std::allocator<char> > > >, std::equal_to<std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >,     std::allocator<std::pair<const std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > >, std::shared_ptr<const Value> >     > > > >::get() const’
   _holder.get<ReturnType>();
           ^
In file included from Value/Value.hpp:8:0,
             from Value/Value.cpp:1:
deps/variant/variant.hpp:680:23: note: candidate: template<class T, typename     std::enable_if<(mapbox::util::detail::direct_type<T, std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > >,     unsigned int, float, bool, std::shared_ptr<const     std::vector<std::shared_ptr<const Value>, std::allocator<std::shared_ptr<const     Value> > > >, std::shared_ptr<const std::unordered_map<std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > >,     std::shared_ptr<const Value>, std::hash<std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >,     std::equal_to<std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > > >,     std::allocator<std::pair<const std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > >, std::shared_ptr<const Value> >     > > > >::index != mapbox::util::detail::invalid_value), void>::type* <anonymous>     > T& mapbox::util::variant<Types>::get() [with T = T; typename     std::enable_if<(mapbox::util::detail::direct_type<T, Types ...>::index !=     mapbox::util::detail::invalid_value)>::type* <anonymous> = <enumerator>; Types =     {std::shared_ptr<const std::basic_string<char, std::char_traits<char>,     std::allocator<char> > >, unsigned int, float, bool, std::shared_ptr<const     std::vector<std::shared_ptr<const Value>, std::allocator<std::shared_ptr<const     Value> > > >, std::shared_ptr<const std::unordered_map<std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > >,     std::shared_ptr<const Value>, std::hash<std::shared_ptr<const     std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >,     std::equal_to<std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > > >,     std::allocator<std::pair<const std::shared_ptr<const std::basic_string<char,     std::char_traits<char>, std::allocator<char> > >, std::shared_ptr<const Value> >     > > >}]
     VARIANT_INLINE T& get()

编译器错误

row*column + offset + size*parameter

Answer 1

这个问题可以关闭，问题是由于我的一个愚蠢的错误，int实际上并不是一个有效的类型，因此模板扣除失败了。

使用decltype作为模板参数

1 个答案: