C++ imperfect(七) -- an ambiguity in a template argument

Template arguments   ( ISO/IEC 14882:2011(E)  § 14.3)                                                                    [temp.arg]
  2  In a template-argument, an ambiguity between a type-id and an expression is resolved to a type-id, regardless  of the form of the corresponding  template-parameter.
[ Example:

// template_argument.cpp
//
#include <iostream>
using namespace std;

template<class T> void f()
{
    cout<<"[template void f()] called\n"<<endl;
}

template<int I> void f()
{
    cout<<"[template void f()] called\n"<<endl;
}

class Functor
{
public:
    int operator()(void)
    {
        cout<<"[int Functor::operator()(void)] called\n"<<endl;
        return true;
    }
};

int g()
{
    return true;
}

int main(int argc, char* argv[])
{
    cout<<"running [f();]"<<endl;
    f<int()>(); // int() is a type-id: call the firt template<class T> void f()

    cout<<"running [f();]"<<endl;
    f<Functor()>();

    cout<<"running [f();]"<<endl;
    f<int>();

    cout<<"running [f<1>();]"<<endl;
    f<1>();

    cout.flush();
    return 0;
}

Result:

running [f<int()>();]
[template<class T> void f()] called

running [f<Functor()>();]
[template<class T> void f()] called

running [f<int>();]
[template<class T> void f()] called

running [f<1>();]
[template<int I> void f()] called

— end example ]
     There  is  no  such  ambiguity  in  a  default  template-argument  because  the  form  of  the  template-parameter  determines  the allowable forms of the  template-argument.