What is the reason for the second brackets <> in the following function template:
template<> void doh::operator()<>(int i)
T
I've looked it up, and found that it is specified by 14.5.2/2:
A local class shall not have member templates. Access control rules (clause 11) apply to member template names. A destructor shall not be a member template. A normal (non-template) member function with a given name and type and a member function template of the same name, which could be used to generate a specialization of the same type, can both be declared in a class. When both exist, a use of that name and type refers to the non-template member unless an explicit template argument list is supplied.
And it provides an example:
template <class T> struct A {
void f(int);
template <class T2> void f(T2);
};
template <> void A<int>::f(int) { } // non-template member
template <> template <> void A<int>::f<>(int) { } // template member
int main()
{
A<char> ac;
ac.f(1); //non-template
ac.f(’c’); //template
ac.f<>(1); //template
}
Note that in Standard terms, specialization
refers to the function you write using an explicit specialization and to the function generated using instantiation, in which case we have to do with a generated specialization. specialization
does not only refer to functions you create using explicitly specializing a template, for which it is often only used.
Conclusion: GCC gets it wrong. Comeau, with which i also tested the code, gets it right and issues a diagnostic:
"ComeauTest.c"
, line 16: error:"void doh::operator()(bool)"
is not an entity that can be explicitly specializedtemplate<> void doh::operator()(bool i)
Note that it isn't complaining about the specialization of the template for int
(only for bool
), since it doesn't refer to the same name and type: The function type that specialization would have is void(int)
, which is distinct from the function type of the non-template member function, which is void(bool)
.