通过函数指针的C ++函数方法

Question

I'm trying to understand how function pointer works can't clarified why I get the following err.

#include <iostream>

using namespace std;
enum COLOR {RED,BLACK,WHITE};

class Car
{public:
 Car (COLOR c): color(c) { cout<<"Car's constructor..."<<endl; CarsNum++;}
 ~Car () {cout<<"Car's destructor..."<<endl; CarsNum--;}
 void GetColor () { cout<<"Color of the car is"<<color<<endl;}
 static int GetCarsNum () {cout<<"Static membet CarsNum="<<CarsNum<<endl; return 0;}
private:
COLOR color;
static int CarsNum;

};

int Car::CarsNum=0;

int main()
{
int (Car::*pfunc) () = NULL;
pfunc=&Car::GetCarsNum ();


 Car *ptr= new Car(RED);
 ptr->GetColor ();
 ptr->GetCarsNum ();
 delete ptr;
 ptr=0;
 Car::GetCarsNum();
    return 0;
}

Err msg:

main.cpp|23|error: lvalue required as unary '&' operand

Problem is with:

  pfunc=&Car::GetCarsNum ();

Any help would be greatly appreciated

Answer 1

With &Car::GetCarsNum () you are calling GetCastNum , and taking the return value to make it a pointer (with the address-of operator & ).

To solve this simply drop the parentheses:

pfunc=&Car::GetCarsNum;

Answer 2

No need for parentness:

pfunc=&Car::GetCarsNum;

and

int (Car::*pfunc) () = NULL;

Oh, UPDATE: you have static method: In this case GetCarsNum() is just a simple function:

int (*pfunc) () = &Car::GetCarsNum;

Answer 3

I think that by putting braces behind the method, you call the method. This confuses the compiler. who now thinks the & operator is used as binary operator. So remove the () and only specify the function name.

Answer 4

Thanks, guys. Now I see the diffrence between pointers to static method (in this case we use syntaxis as for simple function)

int (*pfunc) () = &Car::GetCarsNum;

int this case both with '&' and without '&' gives the same result.(and what is the difference). and calling it:

pfunc();

And other side - function pointer on standart method:

void (Car::*pfunc2) () = NULL;
pfunc2=&Car::GetColor;

and calling it:

(ptr->*pfunc2)();