Hi, I've looked over a few similar questions, but I'm still confused. I'm trying to figure out how to explicitly (not by compiler optimization etc) and C++03-compatible avoid copying of an object when passing it to a template function. Here is my test code:
#include <iostream>
using namespace std;
struct C
{
C() { cout << "C()" << endl; }
C(const C&) { cout << "C(C)" << endl; }
~C() { cout << "~C()" << endl; }
};
template<class T> void f(T) { cout << "f<T>" << endl; }
template<> void f(C c) { cout << "f<C>" << endl; } // (1)
template<> void f(const C& c) { cout << "f<C&>" << endl; } // (2)
int main()
{
C c;
f(c);
return 0;
}
(1) accepts the object of type C, and makes a copy. Here is the output:
C()
C(C)
f<C>
~C()
~C()
So I've tried to specialize with a const C& parameter (2) to avoid this, but this simply doesn't work (apparently the reason is explained in this question).
Well, I could "pass by pointer", but that's kind of ugly. So is there some trick that would allow to do that somehow nicely?
EDIT: Oh, probably I wasn't clear. I already have a templated function
template<class T> void f(T) {...}
But now I want to specialize this function to accept a const& to another object:
template<> void f(const SpecificObject&) {...}
But it only gets called if I define it as
template<> void f(SpecificObject) {...}
