Pointer to template function which is NOT member of any class

Question

I would like to have pointer to template function which has 2 parameters of type T .

template <typename T>
typedef bool( * f )( T, T );

template <typename T>
bool mniejsze (T pierwszy , T drugi){
    if( pierwszy < drugi)
        return true;
    return false;
}
template <typename T>
bool wieksze (T pierwszy, T drugi){
    if( pierwszy > drugi )
        return true;
    return false;
}

But I get:

 error: template declaration of 'typedef'|

EDIT: Then I would like to pass that pointer: Is it the right way?

template <typename T>
T minmax(T a[], int n,bool &f){
    return f(a[0],a[1]);

}

Answer 1

In C++11 you can use aliases:

template<typename T>
using f = bool( *)( T, T );

usage:

f<int> f1 = wieksze;
f1( 3, 4);

http://ideone.com/KyUjwP

In C++03 there is a workoround:

template<typename T>
struct f {
    typedef bool( *type)( T, T );
};

usage:

f<int>::type  f1 = mniejsze<int>;
f<int>::type  f2 = mniejsze<int>;
f1( 3, 4);

template<typename T>
T minmax(T a[], int n, typename f<T>::type fp ){ 
    if ( fp( a[0], a[1])) return 1;
    return -1;
}

int main() {
    // your code goes here
    f<int>::type  f1 = wieksze<int>;
    bool b = f1( 3, 4);
    int a[] = { 3, 4};
    std::cout << minmax<int>( a, 0, f1);
    return 0;
}

http://ideone.com/Dh2eEN

Answer 2

In C++11 you can do:

template<class T>
using f = bool(*)(T, T);

And use:

f<int> fp = wieksze;

Answer 3

How about the following?

 template <typename T>
  struct FNPTR {
    typedef bool (*f)( T, T );
 };

Put a dummy wrapper of a struct, and use FNPTR::f later? For example,

FNPTR::f = mniejsze<char>;

This should work!