templates of functions

Question

I'm told to create template of function , that will take 4 arguments :

• pointer
• reference
• pointer to array
• pointer to function

How to perform this task ? I was trying :

#include <iostream>
using namespace std;

int nothing(int a)
{
    return a;
}

template<typename T> T func(int *L, int &M, char *K, int (*P)(int))
{
    cout << L << "," << M << "," << K[0] << "," << P() << endl;
    return 0;    
}

int main()
{
    int x = 3;
    int *z = &x;
    int &y = x;
    char c[3];
    int (*pf)(int) = nothing;

    cout << "some result of func" << func(z, y, c, pf) << endl;

    system("pause");
    return 0;
}

This gives me "no matching function , I guess for 'pf'. Also now I have no control over what to pass within pf or am I wrong ?

Answer 1

To help you little bit more, try to remember how the "main" function taking arguments looks like, this will help you to see how you can make a pointer to an array.

Answer 2

You now have some problems left...

TYPE (*P)(x) says you expect a pointer to function that takes an argument of type x - change it to an existing type.

In the expression func(z, y, c, pf(x)) you try to call the function pointer pf instead of just passing it.

Then you are calling func with parameters based on different types for the first 3 parameters, int and char, but func expects them to be based on the same type.
Try writing down with what types func will be called with and try matching that to a signature for func with TYPE being substituted to say int.

E.g. if you have the following:

template<typename T> void f(T* a, T* b);

and try to call it like this:

int* a = 0;
int* b = 0;
f(a, b);

the compiler instantiates and calls a function

void f<int>(int*, int*);

But if you do the following:

int*  a = 0;
char* b = 0;
f(a, b);

what should be called?

void f<int> (int*,  int* ); // doesn't match, 2nd argument is char*
void f<char>(char*, char*); // doesn't match, 1st argument is int*