Why does this pointer to C++ function code generate a compile error?

Question

Can anyone solve this? I can’t seem to find the solution anywhere, but I see no logical reason why the line below (with the comment showing the compile error) should be a problem.

Note: This question is a derivative of How can a C++ base class determine at runtime if a method has been overridden?

class MyClass
{
        typedef void (MyClass::*MethodPtr)();  


        virtual void Method()
        {
                MethodPtr a = &MyClass::Method; // legal
                MethodPtr b = &Method;  // error C2276: ‘&’ : illegal operation on bound member function expression

                if (a == b)     // this method has not been overridden?
                        throw “Not overridden”;
        }
};

Vijay Mathew · Accepted Answer · 2009-11-26 08:06:16Z

12

ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. This takes care of name mangling. So what you are trying to do will not work in a standards compliant C++ compiler.

edited Nov 26, 2009 at 8:06

answered Nov 26, 2009 at 7:28

Vijay Mathew

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1 Comment

Conrad Meyer Over a year ago

If you could do this, C++ would have something like [closures][1]. [1]: en.wikipedia.org/wiki/Closure_%28computer_science%29

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