C++ Syntax to call a member function using it's function pointer

Question

The example below uses a function pointer to a member function of the class Blah. The syntax of the function pointer is clear to me. However when calling I had to put brackets around this->*funcPtr and I am not sure why this is required. I guess it is related to the way how C++ evaluates the expression. The compiler used is VS 2008.

#include <iostream>

using namespace std;

struct Blah {

    void myMethod(int i, int k) {
        cout << "Hi from myMethod. Arguments: " << i << " " << k << endl;
    }

    typedef void (Blah::*blahFuncPtr)(int, int);

    void travelSomething(blahFuncPtr funcPtr) {
        (this->*funcPtr)(1, 2);
        // w/o the brackets I get C2064 in VS 2008
        // this->*funcPtr(1, 2);
    }
};

int main() {
    Blah blah;
    blah.travelSomething(&Blah::myMethod);
    cin.get();
    return 0;
}

parashift.com/c++-faq-lite/pointers-to-members.html#faq-33.6 — Mat
– Mat, Commented Feb 27, 2012 at 6:54
Ok, maybe a macro would be a good idea. But that still doesn't answer my question. I would like to know why.. — Nils
– Nils, Commented Feb 27, 2012 at 6:56
Uh and how would one define a macro for pointers to function which take arguments using variadic macros for example? — Nils
– Nils, Commented Feb 27, 2012 at 7:06

jogojapan · Accepted Answer · 2012-02-27 07:01:03Z

2

The function call operator () takes higher precendence than the 'pointer to member' operator ->*.

See, for example, here.

answered Feb 27, 2012 at 7:01

jogojapan

70.4k11 gold badges110 silver badges136 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

C++ Syntax to call a member function using it's function pointer

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related