如何在编译时替换元组元素？

Question

有没有办法在编译时替换元组元素？

例如，

using a_t = std::tuple<std::string,unsigned>;  // start with some n-tuple
using b_t = element_replace<a_t,1,double>;     // std::tuple<std::string,double>
using c_t = element_replace<b_t,0,char>;       // std::tuple<char,double>

Answer 1

你可以用这个：

// the usual helpers (BTW: I wish these would be standardized!!)
template< std::size_t... Ns >
struct indices
{
    typedef indices< Ns..., sizeof...( Ns ) > next;
};

template< std::size_t N >
struct make_indices
{
    typedef typename make_indices< N - 1 >::type::next type;
};

template<>
struct make_indices< 0 >
{
    typedef indices<> type;
};

// and now we use them
template< typename Tuple, std::size_t N, typename T,
          typename Indices = typename make_indices< std::tuple_size< Tuple >::value >::type >
struct element_replace;

template< typename... Ts, std::size_t N, typename T, std::size_t... Ns >
struct element_replace< std::tuple< Ts... >, N, T, indices< Ns... > >
{
    typedef std::tuple< typename std::conditional< Ns == N, T, Ts >::type... > type;
};

然后像这样使用它：

using a_t = std::tuple<std::string,unsigned>;     // start with some n-tuple
using b_t = element_replace<a_t,1,double>::type;  // std::tuple<std::string,double>
using c_t = element_replace<b_t,0,char>::type;    // std::tuple<char,double>

Answer 2

看看boost MPL transform或replace algos http://www.boost.org/doc/libs/1_40_0/libs/mpl/doc/refmanual/transformation-algorithms.html

Answer 3

您可以使用std :: tuple_element访问元组类型的元素类型。 这实际上不允许您替换元组元素类型，但它允许您根据在其他元组类型中用作元素类型的类型来定义元组类型。