问题
Explanation :
std::rank just works for c style array .
So I implemented similar rank
for std::vector
which works fine :
#include <iostream>
#include <vector>
template<typename Type, Type val>
struct integral_constant
{
static constexpr Type value =val;
};
template<typename>
struct rank
: public integral_constant<std::size_t, 0> { };
template<typename Type>
struct rank< std::vector<Type> >
: public integral_constant<std::size_t, 1 + rank<Type>::value> { };
template<class T>
constexpr size_t vector_dimentions(T)
{
return rank<T>::value ;
}
int main()
{
std::vector<std::vector<std::vector<int>>> vec;
std::cout<<vector_dimentions(vec) << '\n';
}
ideone
Problem :
Now I want to generalize it for other containers like std::list
,...
So I change the struct definition to :
template<template<typename>class Container,typename Type>
struct rank< Container<Type> >
: public integral_constant<std::size_t, 1 + rank<Type>::value> { };
ideone
But now It gives wrong answer(always 0) !
I think in this case it can't deduce the right struct because it has now 2 template parameter . is it correct ?! How can I solve that ?
回答1:
With help of KerrekSB I found the solution :
template <typename> struct prank : std::integral_constant<std::size_t, 0> {};
template <template <typename...> class C, typename ...Args>
struct prank<C<Args...>>
: std::integral_constant<
std::size_t,
1 + prank<typename C<Args...>::value_type>::value> {};
template <typename U, typename V>
struct prank<std::pair<U, V>>
: std::integral_constant<std::size_t, 1 + prank<V>::value> {};
template <typename... Args>
struct prank<std::tuple<Args...>>
: std::integral_constant<std::size_t, 1> {};
template <typename T,typename... Args>
struct prank<std::tuple<T,Args...>>
: std::integral_constant<std::size_t, prank<T>::value+prank<std::tuple<Args...>>::value> {};
ideone
来源:https://stackoverflow.com/questions/24222427/implementing-stdrank-for-other-containers