这不是 secret ,std::get<i>(tuple)
惹恼了许多程序员。
我想使用类似 tuple[i]
的东西来代替它.
所以我尝试模拟它。
#include <iostream>
#include <type_traits>
#include <tuple>
template < int > struct index{};
template< char ... >
struct combine;
template<> struct combine<> : std::integral_constant< int , 0>{};
constexpr int ten(size_t p)noexcept
{
return p == 0 ? 1 : 10 * ten(p-1);
}
template< char c, char ... t>
struct combine<c, t...> : std::integral_constant< int, (c - '0')*ten(sizeof...(t)) + combine<t...>::value >
{ static_assert(c >= '0' && c <= '9', "only 0..9 digits are allowed"); };
template< char ... c >
constexpr auto operator "" _index()noexcept
{
return index< combine<c...>::value >{};
};
template< class ... Args >
struct mytuple : public std::tuple<Args...>
{
using std::tuple<Args...>::tuple;
template< int i >
auto& operator []( index<i> ) noexcept
{
return std::get< i > ( static_cast< std::tuple<Args...> & >(*this) );
}
template< int i>
auto const& operator [](index<i> )const noexcept
{
return std::get< i >(static_cast< std::tuple<Args...> const& >(*this) );
}
};
int main()
{
static_assert( combine<'1','2','3','4'>::value == 1234, "!");
static_assert( std::is_same< index<785>, decltype( 785_index ) > {}, "!");
using person = mytuple< std::string, int, double, char>;
person s = std::make_tuple("Bjarne Stroustrup", 63, 3.14, '7' );
auto name = s[ 0_index ];
auto old = s[ 1_index ];
auto number = s[ 2_index ];
auto symbol = s[ 3_index ];
std::cout << "name: " << name << '\t'
<< "old: " << old << '\t'
<< "number: " << number<< '\t'
<< "symbol: " << symbol<< '\t'
<< std::endl;
}
问:这段代码有什么问题?即这个代码是否可用?
如果可用为什么不可用 std::tuple
像这样实现?
最佳答案
我不太确定您的情况的具体问题是什么,但您似乎正在寻找一点方便。下面使用占位符(以 _1
开头)来简化您的代码:
#include <iostream>
#include <type_traits>
#include <tuple>
#include <functional>
template< class ... Args >
struct mytuple : public std::tuple<Args...>
{
using std::tuple<Args...>::tuple;
template< typename T >
auto& operator []( T ) noexcept
{
return std::get< std::is_placeholder<T>::value - 1 >( *this );
}
template< typename T >
auto const& operator []( T ) const noexcept
{
return std::get< std::is_placeholder<T>::value - 1 >( *this );
}
};
int main()
{
using person = mytuple< std::string, int, double, char>;
using namespace std::placeholders;
person s = std::make_tuple("Bjarne Stroustrup", 63, 3.14, '7' );
auto name = s[ _1 ];
auto old = s[ _2 ];
auto number = s[ _3 ];
auto symbol = s[ _4 ];
std::cout << "name: " << name << '\t'
<< "old: " << old << '\t'
<< "number: " << number<< '\t'
<< "symbol: " << symbol<< '\t'
<< std::endl;
}
这里的技巧是要知道 std::is_placeholder<T>
保证源自std::integral_constant<int,N>
:) 希望你喜欢它。
关于c++ - 通过索引访问元组元素 c++11,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19293382/