Thread: Template function pointer???

I am learning c++ and in the past two week, I started learning and practicing about Function Pointers and Templates.. but it came to my mind if I can have a template function pointer..
So, here is what I have coded as I started my practice to try to understand whats going on... but Visual Studio 2013 is not accepting my function call in the main method... not sure what I am doing wrong and its saying that my function pointer is undefined but I thought it was?

Code:

#include <iostream>

template<class T>
int foo(T& x){
    return x * 500;
}

template<class T>
T (*myFunc)(T) = foo;

int main(){

    std::cout << myFunc(7) << std::endl;

    system("pause");
    return 0;
}

Well, I really hope you guys can help me to understand if what I want to do is possible and which is the proper way and why?
Thanks a lot.. if required more details my questions please let me know!!!

Even if it were to work, you've got some mismatches (for instance, myFunc<int> (if it were to exist) would accept an int parameter, but foo<int> accepts an int& parameter).

I suspect that function pointers can't be templated, but I'm not in front of my standard to check. The following does work:

Code:

#include <iostream>

template <class T>
int foo(T x) {
    return x * 500;
}

int (*myFunc)(int) = foo<int>;

int main() {

    std::cout << myFunc(7) << std::endl;
    return 0;
}

Note I made foo accept a T instead of a T& (otherwise you can't pass a const literal like "7" to it), and the function pointer is of a definite type.

Originally Posted by tabstop

Even if it were to work, you've got some mismatches (for instance, myFunc<int> (if it were to exist) would accept an int parameter, but foo<int> accepts an int& parameter).

I suspect that function pointers can't be templated, but I'm not in front of my standard to check. The following does work:

Code:

#include <iostream>

template <class T>
int foo(T x) {
    return x * 500;
}

int (*myFunc)(int) = foo<int>;

int main() {

    std::cout << myFunc(7) << std::endl;
    return 0;
}

Note I made foo accept a T instead of a T& (otherwise you can't pass a const literal like "7" to it), and the function pointer is of a definite type.

Thanks for the help, I was actually working within the program to also try the answer you gave me.. I had the mismatch of T& in step of T..

Thanks, no.. It was just a question that came up to my head and I was looking for answer.. I am just practicing some new topics learned and just thought if I could do something like that..