Nice way of explaining the Function pointers.

I tried to compile the code given up and I am still getting this error

error: prototype for ‘void Kas::secondfunction(int (Kas::*)(int), int)’ does not match any in class ‘Kas’
error: candidate is: void Kas::secondfunction(int (*)(int), int)

void Kas::secondfunction ( int (Kas::*ptofunction)(int),int t){
	 (this->*ptofunction)(t);
};

void Kas::thirdfunction (){
	 secondfunction(&Kas::firstfunction,3);
};

I write a class.

There is a function, member of this class.

There is another function, member of this class, takes a “function pointer” as parameter.

Inside this class there is also a third function and it calls the second function with a function pointer to first function as parameter.

Here is the code:

#include <iostream>
using namespace std;

class Kas {

	 public:
	  int ada;

	  int firstfunction (int);

	  void secondfunction (int (*ptofunction)(int), int t);

	  void thirdfunction ();

};

void Kas::secondfunction ( int (*ptofunction)(int),int t){
	 (*ptofunction)(t);
};

int Kas::firstfunction (int s) {
	 return ada+s;
};

void Kas::thirdfunction (){
	 secondfunction(firstfunction,3);	//The problem is here!!
};

int main () {

    Kas den;
    den.thirdfunction();
    cout<<den.ada<<"\n";

    system("PAUSE");
    return 0;

}

how can I call the second function?

Now I get it!!!

Probably forget again tomorrow, but at least I can come here and look it up again!

THANK YOU!!!!

1. You forgot to include iostream needed to define the ‘cout’

2.

void PrintArray(int *pArray, int nSize)
{
    for (int iii=0; iii < nSize; iii++)
        cout << pArray[iii] << " ";
    cout << endl;
}

You forgot to add ‘using namespace std;’ thus cout doesn’t work here.

1.int foo(int nX)
2.{
3.}
4.
5.int (*pFoo)(int) = foo; // assign pFoo to foo()
6.
7.(*pFoo)(nValue); // call function foo(nValue) through pFoo.

The second way is via implicit dereference:
view source
print?
1.int foo(int nX)
2.{
3.}
4.
5.int (*pFoo)(int) = foo; // assign pFoo to foo()
6.
7.pFoo(nValue); // call function foo(nValue) through pFoo.

Thanks for catching this.