テンプレートの推論された型をお手軽確認 ref: http://qiita.com/_EnumHack/items/7c8556ab6ddba3a785f8

file0.cpp

template < typename T >
void f(T&&){}

void g(int const&){};

int main(){
    f(g);
}

file1.cpp

#include <iostream>
#include <typeinfo>
using namespace std;

template < typename T >
void f(T&&){ cout << typeid(T).name() << endl;}

void g(int const&){};

int main(){
    f(g);
}

file2.cpp

#include <iostream>
#include <typeinfo>
#include <cxxabi.h>
using namespace std;

template < typename T >
void f(T&&){
    const type_info& id = typeid(T);
    int stat;
    char *name = abi::__cxa_demangle(id.name(),0,0,&stat);
    if( name!=NULL ) {
        if( stat==0 ) {    // ステータスが0なら成功
            printf("T = %s",name);
        }
        free(name);        // freeする必要がある
    }
}

void g(int const&){};

int main(){
    f(g);
}

file3.cpp

#include <iostream>
#include <boost/type_index.hpp>
using namespace std;

template < typename T >
void f(T&&){ cout << boost::typeindex::type_id_with_cvr<T>().pretty_name() << endl;}

void g(int const&){};

int main(){
    f(g);
}

file4.cpp

template < typename T >
[[deprecated]] void f(T&&){}

void g(int const&){};

int main(){
    f(g);
}

file5.txt

prog.cc: In function 'int main()':
prog.cc:8:8: warning: 'void f(T&&) [with T = void (&)(const int&)]' is deprecated [-Wdeprecated-declarations]
     f(g);
        ^

>prog.cc:3:21: note: declared here
 [[deprecated]] void f(T&&){}
                     ^