问题
I cant see why the statement in main is ambiguous.
template<class T, class U, int I> struct X
{ void f() { cout << "Primary template" << endl; } };
template<class T, int I> struct X<T, T*, I>
{void f() { cout << "Partial specialization 1" << endl;}};
template<class T, class U, int I> struct X<T*, U, I>
{void f() { cout << "Partial specialization 2" << endl;}};
template<class T> struct X<int, T*, 10>
{void f() { cout << "Partial specialization 3" << endl;}};
template<class T, class U, int I> struct X<T, U*, I>
{void f() { cout << "Partial specialization 4" << endl;}};
int main()
{
X<int, int*, 10> f;
}
Isn't X<int, T*, 10> the most specialized template?
This is an example from http://publib.boulder.ibm.com/infocenter/lnxpcomp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8l.doc%2Flanguage%2Fref%2Fpartial_specialization.htm
回答1:
A template specialization is more specialized than another if every argument list that matches the first also matches the second, but not the other way around.
When looking at X<int, T*, 10> and X<T, T*, I>:
X<int, float*, 10>matches the first but not the second.X<float, float*, 10>matches the second but not the first.
Therefore neither is more specialized than the other, and a template instantiation that matches both specializations won't compile.
来源：https://stackoverflow.com/questions/8637145/partial-template-specialization-ambiguity