Function pointers, function pointers, and pointer Functions

Function pointers, function pointers, and pointer Functions
(1/2) Type

A normal function pointer cannot be assigned the address of a member function.

Int (* pFunc )();

PFunc is a function pointer, while int (*) () is a type.

The member function address must be assigned to the member function pointer.

class Base{public:    int func() { return 1; }};int main(){    int (*pFunc)();    pFunc = &Base::func;  //error: cannot convert 'int (Base::*)()' to 'int (*)()' in assignment    return 0;}

class Base{public:    int func() { return 1; }};int main(){    int (Base::*pFunc)(); // OK    pFunc = &Base::func;    return 0;}

Again:

class Base{public:    int func() { return 1; }};class Derived : public Base{public:    int foo() { return 1; }};int main(){    typedef int (Base::*FuncPtr)();    FuncPtr fPtr = &Derived::foo; // error    return 0;}

Mandatory type conversion required

......

FuncPtr fPtr = (FuncPtr)&Derived::foo;  // OK

In turn, the base class member function address can be assigned a pointer to the member function of the derived class.

class Base{public:    int func() { return 1; }};class Derived : public Base{public:    int foo() { return 1; }};int main(){    typedef int (Derived::*FuncPtr)();    FuncPtr fPtr = &Base::func;  // OK    return 0;}

(2/2) Call

class Base{public:    virtual void func(int i) { cout << "Base: " << i << endl; }};class Derived : public Base{public:    virtual void func(int i) { cout << "Derived: " << i << endl; }};int main(){    typedef void (Base::*FuncPtr)(int);    FuncPtr fPtr = &Base::func;    Base b;    (b.*fPtr)(2);    Derived* dPtr = new Derived();    (dPtr->*fPtr)(2);    Base* bPtr = new Derived();    (bPtr->*fPtr)(2);    return 0;}

Output:

Base: 2
Derived: 2
Derived: 2