INFO: Creating a Function Pointer to a C++ Member Function (94579)

MORE INFORMATION

In the C and C++ languages, an application can define a variable that contains the address of a function. You can call the function using this variable instead of through the function name. In C++, a pointer to a nonstatic member function contains the address of the function in the class, not in the object. To call the function, use the appropriate member selection operator (. or ->), the indirection operator (*), and the name of an object of the class.

In most cases in the C language, function pointer declarations take the following form:

   int (*ptr)();
				

This code declares a pointer to a function that returns an integer. The function accepts an unknown number of augments. To create a pointer to a C++ class member function, specify the class name in the function pointer declaration, as follows:

   int (Sample::*ptr)();
				

This code declares a pointer to a member function in the Sample class that returns an integer. This function does not accept any arguments.

The different ways to interpret the empty parentheses in function declarations is a major difference between the C and C++ languages. In C, a function pointer declaration that has no arguments is syntactically identical to the following:

   int (*ptr)(...);
				

However, a C++ prototype declared without arguments is syntactically identical to the following:

   int (Sample::*ptr)(void);
				

In C++, if the function accepts arguments, the types of the arguments must also be listed, as in the following:

   int (Sample::*ptr2)(int, int);
				

In C, the types of the arguments may be listed in the same manner. A pointer to a static member function is declared in the same manner as a pointer to a C function. However, because the declaration is part of a C++ program, the declaration must list any arguments and their associated types.

To determine the procedure address to assign to a pointer variable, use the class name and the scope resolution operator (::). This syntax provides flexibility because a declared variable can contain the address of any object of the class. The object name in the function call determines the copy of the function used.

The sample code below demonstrates declaring and using pointers to a class member function and to a static member function. Note that when the arguments in the function pointer declaration do not match the arguments of the function assigned to the pointer, the compiler generates errors C2440 and C2564. For example, if the "int" declaration is omitted from the argument list declaration for the function pointer, the compiler generates the following error messages:

16-bit

32-bit

Sample Code

/*
 * Compile options needed: None
 */ 

#include <iostream.h>

class Data
{
private:
   int y;
   static int x;

public:
   void SetData(int value) {y = value; return;};
   int GetData() {return y;};
   static void SSetData(int value) {x = value; return;};
   static int SGetData() {return x;};
};

int Data::x = 0;

void main(void)
{
   Data mydata, mydata2;

   // Initialize pointer.
   void (Data::*pmfnP)(int) = &Data::SetData; // mydata.SetData;

   // Initialize static pointer.
   void (*psfnP)(int) = &Data::SSetData;

   mydata.SetData(5); // Set initial value for private data.
   cout << "mydata.data = " << mydata.GetData() << endl;

   (mydata.*pmfnP)(20); // Call member function through pointer.
   cout << "mydata.data = " << mydata.GetData() << endl;

   (mydata2.*pmfnP)(10) ; // Call member function through pointer.
   cout << "mydata2.data = " << mydata2.GetData() << endl;

   (*psfnP)(30) ; // Call static member function through pointer.
   cout << "static data = " << Data::SGetData() << endl ;
}
				

For more information about parameter lists in function declarations and how these are handled by C and C++, please see the following article in the Microsoft Knowledge Base: