I want to declare an ordinary pointer (not a member pointer) to a member function. I have tried but apparently I am getting errors. Is this allowed in C++? Why is it not?
#include<iostream>
class A{
private:
int m;
public:
void show(void){
std::cout<<"M= "<<m<<std::endl;
}
};
int main()
{
A a;
void (*ptr)(void);
ptr = a.show;
*ptr;
return 0;
}
The error I am getting is: argument of type ‘void (A::)()’ does not match ‘void (*)()’
You can't get a non-member pointer to a member function. If you did, and called it, what would the function use for this?
A pointer to member function still has to be invoked with an object of the appropriate type. You might be able to declare the pointer like this: void (A.*ptr)(void); A.ptr = a.show; but I'm not 100% certain of that. In any case, this would be, IMHO, really crappy code. It violates the KISS principal in so many ways!
Memeber functions are members of a class. Because of that they are not the same as free functions when it comes to function pointers.
void (*ptr)(void);
Will match any function that returns void at takes no parameters that is not a memeber function.
void (A::*ptr)(void)
Will match any function that returns void and takes no parmeters that is a memeber of A.
The reason for this is
void show(void)
Actually has a hidden parameter in it as non static class memeber functions take in the this pointer from the instance that calls it.
Thank you for your replies @gusano79, @rubberman & @NathanOliver!
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.