CODEWITHSUNDEEP
c++pointersfunction-pointerspointer-to-member
q126y
q126y

Reputation: 1671

Why & operator is needed for taking address of member functions but not for global functions?

The following code runs fine

#include <iostream>
using namespace std;

void fun()
{
    cout<<"having some fun";
}

typedef void (*funptr)();
int main() {
    // your code goes here

    funptr p=fun;

    p();
    return 0;
}

This one doesn't work.

#include <iostream>
using namespace std;

class myclass
{
    public:
    void fun()
    {
        cout<<endl<<"having fun inside myclass"<<endl;
    }
};

typedef void (myclass::*funptr)();

int main() {
    // your code goes here

    funptr p=myclass::fun; //THIS DOESN'T WORK. WHY?

    myclass m;

    (m.*p)();
    return 0;
}

Why is & operator needed for member functions?

Upvotes: 0

Views: 101

Answers (1)

TartanLlama
TartanLlama

Reputation: 65730

Lvalues of function type T can be converted to T* implicitly ([conv.func]/4.3), but there is no analogous rule for member functions.

I'm not sure what the reasoning behind this is, so I guess my answer is "the standard says so". Perhaps it's because member function pointers are not used too often, so mandating extra implementation details for them was seen as unnecessary.

Upvotes: 2

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