Quiz : function pointers as template arguments

Yesterday, I came across a piece of template code that took me a little by surprise (because I had not come across something like this before) but I was able to put my reasoning through. I will share the code first:

[CODE]

#include <iostream>

template<typename T>
void foo(const T&)
{
               std::cout << "const";
}

template<typename T>
void foo(T&)
{
               std::cout << "non-const";
}

void bar() { }

int main()
{
               foo(bar);
}

The question was - what would the program print? Will the argument "bar" resolve as a parameter to the first template having argument type "const F&" or to the second template having non-const argument type "F&"?

The easiest way to check for the resolution is to ask for explicit template instantiation of the "foo" function template. How? Here is how:

[CODE]

int main()
{
               typedef void (*f_ptr)(); //create a typedef for functions like bar
                                                               //taking no arguments and having return type as void.
               foo<const f_ptr>(bar); //1
               foo<f_ptr>(bar); //2
}

After that, just remove one of the "foo" templates. So, the code to check for compilation is this:

[CODE]
#include<iostream>

template<typename T>
void foo(T&)
{
               std::cout << "non-const";
}

void bar() { }

int main()
{
               typedef void (*f_ptr)(); //create a typedef for functions like bar
                                                               //taking no arguments and having return type as void.
               foo<const f_ptr>(bar); //1
               foo<f_ptr>(bar); //2
}


If you removed the second template, code compiles fine. But if you kept the second one and removed the first one, you will see that the compilation fails for mis-match in the argument type in statement "//2".

Problem solved. Isn't it? What does this tell about the argument "bar" ? It tells that it is a constant. And hence the call in the initial sample code would resolve to the template instance having the argument type declared const. It is in a way similar to any other type constants, for example 5, 100, 2000 are integral constants, they are literals. And when you declare something as say int i; here i is a variable that can be modified. but 5, 100, or 2000 cannot be. In our initial code, both the templates were capable of instantiating the right function for the argument. In both's presence, the argument match has to be exact and hence instantiation happens from the const one but in its absense the instantiation can happen even with the non-const one as the call can help the template to instantiate over type const f_ptr instead of just f_ptr (which is the case for the const one).

Functions pointers when being passed by taking the address of the function directly is a value of const function pointer type.