#include <iostream>

template<class F> struct function_caller;

template<class R, class P1, class P2> struct function_caller<R (P1, P2)>
{
};

int main()
{
	typedef function_caller<void (int, char)> my_caller;
	std::cout << typeid(my_caller).name() << std::endl;

	return 0;
}

The program output:

struct function_caller<void __cdecl(int,char)>

Where I can read about specialization by the function types (in English) ?

I hope your still waiting for an answer.

First of all the Output of

typeid().name()

is compiler specific, if you didn't know that. For example my compile (gcc 4.1.1) gives me

15function_callerIFvicEE

as output.

I don't know any resources for template specialization specifically for function types.

What you do is a partial template specialization to get your template code to work with function types as template parameters while extracting Informations about return and argument types of the function. And that's a little bit tricky as you can see in your code.

If you have more questions or if I need to describe it more detailed just ask for it

And for your english resources (not everything might seem useful at first):
http://www.parashift.com/c++-faq-lite/index.html
http://www.decadentplace.org.uk/womble/c++/template-faq.html
http://www.tutok.sk/fastgl/callback.html

My .name() is better than yours :p

If you have more questions or if I need to describe it more detailed just ask for it

Yes! I have questions:

1. Why line 3 in the code is necessary?
2. Can I achieve the same result (passing a function type into template) without using the specialization technique?

IMHO.

SAn schrieb:

Can I achieve the same result (passing a function type into template) without using the specialization technique?

yes, you can:

#include <iostream>

template<class F> struct function_caller
{
};

int main()
{
    typedef function_caller<void (int, char)> my_caller;
    std::cout << typeid(my_caller).name() << std::endl;

    return 0;
}

the first declaration of function_caller only tells the compiler that it has one template parameter. and void (int,char) is not a compound, but one distinct type (which is used as a template argument in this example)

...the first declaration of function_caller only tells the compiler that it has one template parameter. and void (int,char) is not a compound, but one distinct type (which is used as a template argument in this example)

Hmm...
Thank you.