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;
}

Answer 1

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

See, for example, here.