c++ - 按索引访问元组元素 c++11

Question

这不是 secret ，std::get<i>(tuple)惹恼了许多程序员。

取而代之，我想使用类似 tuple[i] 的东西.

所以我试着模拟了一下。

#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？即此代码是否可用？ 如果可用，为什么不可用 std::tuple是这样实现的？

Answer 1

我不太确定您的具体问题是什么，但您似乎正在寻找一些便利。以下使用占位符(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;
}

这里的技巧是知道 std::is_placeholder<T> 保证派生自 std::integral_constant<int,N> :) 希望你喜欢。