2
在下面的代碼中,當模板參數T
的值爲B<m>
時,我嘗試專門化基類A
。模板專業化的不正確實例化
#include <array>
#include <iostream>
template <std::size_t m>
struct B
{
std::array<double,m> arr;
};
template <std::size_t n, typename T>
struct A
{
std::array<T,n> arr;
const static int inner_dim = T::inner_dim;
};
template <std::size_t n >
template <std::size_t m>
struct A<n,B<m>>
{
std::array<B<m>,n> arr;
const static int inner_dim = m;
};
int main(int argc, char *argv[])
{
A<5,A<4,B<3>>> a;
std::cout << a.inner_dim << std::endl;
A<5,B<4>> b;
std::cout << b.inner_dim << std::endl;
return 0;
}
然而,在主要的實例,我收到以下錯誤,當我編譯使用g ++ 5.4:
$ g++ -Wall --std=c++11 specialization.cc
specialization.cc: In instantiation of ‘const int A<4ul, B<3ul> >::inner_dim’:
specialization.cc:15:20: required from ‘const int A<5ul, A<4ul, B<3ul> > >::inner_dim’
specialization.cc:30:18: required from here
specialization.cc:15:20: error: ‘inner_dim’ is not a member of ‘B<3ul>’
const static int inner_dim = T::inner_dim;
^
specialization.cc: In instantiation of ‘const int A<5ul, B<4ul> >::inner_dim’:
specialization.cc:33:18: required from here
specialization.cc:15:20: error: ‘inner_dim’ is not a member of ‘B<4ul>’
看來只能用於一般的定義,從來沒有專業化。我如何確保實現正確的專業化?
注:鐺拒絕嘗試專業化 –