What is the reason for the second brackets <> in the following function template:
template<> void doh::operator()<>(int i)
This came up in SO question where it was suggested that there are brackets missing after operator()
, however I could not find the explanation.
I understand the meaning if it was a type specialization (full specialization) of the form:
template< typename A > struct AA {};
template<> struct AA<int> {}; // hope this is correct, specialize for int
However for function templates:
template< typename A > void f( A );
template< typename A > void f( A* ); // overload of the above for pointers
template<> void f<int>(int); // full specialization for int
Where does this fit into this scenarion?:
template<> void doh::operator()<>(bool b) {}
Example code that seems to work and does not give any warnings/error (gcc 3.3.3 used):
#include <iostream>
using namespace std;
struct doh
{
void operator()(bool b)
{
cout << "operator()(bool b)" << endl;
}
template< typename T > void operator()(T t)
{
cout << "template <typename T> void operator()(T t)" << endl;
}
};
// note can't specialize inline, have to declare outside of the class body
template<> void doh::operator()(int i)
{
cout << "template <> void operator()(int i)" << endl;
}
template<> void doh::operator()(bool b)
{
cout << "template <> void operator()(bool b)" << endl;
}
int main()
{
doh d;
int i;
bool b;
d(b);
d(i);
}
Output:
operator()(bool b)
template <> void operator()(int i)