C ++ - 迭代一个元组和一个元组。类型与常量参数的分辨率
我目前正在为元组编写算术运算符重载。运算符迭代元组以对其每个元素执行操作。这是operator + =的定义：
template< typename... Ts, std::size_t I = 0 >
inline typename std::enable_if< I == sizeof... (Ts), std::tuple< Ts... >& >::type operator +=(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
return lhs;
}
template< typename... Ts, std::size_t I = 0 >
inline typename std::enable_if< I != sizeof... (Ts), std::tuple< Ts... >& >::type operator +=(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
std::get< I >(lhs) += std::get< I >(rhs);
return operator +=< Ts..., I + 1 >(lhs, rhs);
}
std::tuple< int, int, int, int > a = std::make_tuple(1, 2, 3, 4), b = std::make_tuple(5, 6, 7, 8);
a += b;
:/Workspace/raster/main.cpp:833:7: instantiated from here
C:/Workspace/raster/main.cpp:809:45: error: no matching function for call to 'operator+=(std::tuple<int, int, int, int>&, const std::tuple<int, int, int, int>&)'
C:/Workspace/raster/main.cpp:809:45: note: candidates are:
C:/Workspace/raster/main.cpp:800:151: note: template<class ... Ts, unsigned int I> typename std::enable_if<(I == sizeof (Ts ...)), std::tuple<_TElements ...>&>::type operator+=(std::tuple<_TElements ...>&, const std::tuple<_TElements ...>&)
C:/Workspace/raster/main.cpp:806:83: note: template<class ... Ts, unsigned int I> typename std::enable_if<(I != sizeof (Ts ...)), std::tuple<_TElements ...>&>::type operator+=(std::tuple<_TElements ...>&, const std::tuple<_TElements ...>&)
std::enable_if
template< std::size_t I, typename... Ts >
inline typename std::enable_if< I == sizeof... (Ts), std::tuple< Ts... >& >::type assignadd_impl(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
return lhs;
}
template< std::size_t I, typename... Ts >
inline typename std::enable_if< I != sizeof... (Ts), std::tuple< Ts... >& >::type assignadd_impl(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
std::get< I >(lhs) += std::get< I >(rhs);
return assignadd_impl< I + 1, Ts... >(lhs, rhs);
}
template< typename... Ts >
inline std::tuple< Ts... >& operator +=(std::tuple< Ts... >& lhs, const std::tuple< Ts... >& rhs)
{
return assignadd_impl< 0, Ts... >(lhs, rhs);
}
没有找到相关结果
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1 个回复
久纳禾澎
移动到模板参数列表的末尾并相应地更改调用使代码工作。当前C ++ 0x草案的第14.8.2.1节规定，不能从函数调用推导出不在模板参数列表末尾的参数包（这会使原始代码失败），但明确指定所有模板在所有情况下，的参数仍会导致SFINAE错误。上一个问题有一个指向禁止它的确切文本的链接; IBM的文档说这也是一个错误。