Function template specialization format

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