函数模板专业化格式

第二个括号的原因是什么<>在以下函数模板中:

template<> void doh::operator()<>(int i)

SO question 其中建议后面缺少括号 operator()

template< typename A > struct AA {};
template<> struct AA<int> {};         // hope this is correct, specialize for int

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

template<> void doh::operator()<>(bool b) {}

示例代码似乎有效,但没有给出任何警告/错误(使用了gcc 3.3.3):

#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);
}

输出:

operator()(bool b)
template <> void operator()(int i)