Why does the following compile with clang
but not with g++ 4.9
#include <array>
template< typename T1, typename T2 , typename T3 = int>
struct A;
template<typename T, unsigned int N, typename T2, typename T3>
struct A< std::array<T,N>, T2, T3 > {
int a;
};
int main()
{
A< std::array<int,10>, double> a;
a.a +=3;
}
http://coliru.stacked-crooked.com/a/c7800f49ba5aac43
g++ does not find a suitable specialization and complains with "incomplete type". I am wondering since, the default argument typename T3 = int
should apply for the specialization (or does it only apply for full specialization?)
The templates A<T1, T2, T3>
and A<T1, T2>
are not complete defined so you can't use your members, you can solve the problem defining this templates at this way:
#include <array>
template< typename T1, typename T2 , typename T3 = int>
struct A {
int a;
};
template<typename T, unsigned int N, typename T2, typename T3>
struct A< std::array<T,N>, T2, T3 > {
int a;
};
int main()
{
A< std::array<int,10>, double> a;
a.a +=3;
}
Another fine example of specialization more simple:
template<typename T, unsigned int N>
struct A {
T a = N;
};
template<unsigned int N>
struct A<int, N> {
int a = 2*N;
};
#include <iostream>
using namespace std;
main() {
A<float, 30> af;
A<int, 30> ai;
cout << af.a << endl << ai.a << endl;
}
Like @dys say's on your comment using std::size_t
instead unsigned int
works:
template< typename T1, typename T2 , typename T3 = int>
struct A;
template<typename T, std::size_t N, typename T2, typename T3>
struct A< std::array<T,N>, T2, T3 > {
T3 a = N;
int b;
};
int main()
{
A< std::array<int,10>, double> a;
a.a +=3;
A< std::array<int,10>, double, int> b;
b.b +=3;
}
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