Question

I have my code organized as following

class MyClass
{
public:
   typedef void (MyClass::Ptr2func)();
}

struct abc
{
    MyClass::Ptr2func ptr;
    bool result;
}

void main()
{
    MyClass myCls;
    abc  var;
    //here is a lot of decision making code and pointer initializations

    //I want to copy the pointer to function in another variable
    MyClass::Ptr2func tempPtr;
    tempPtr=var.ptr;
}

When I try to copy the var.ptr into tempPtr it gives me a compilation error that the argument list is missing. Also it gives me compilation error on myCls.*(var.ptr()); Is there a precedence issue? I have tried using parenthesis but nothing works. I hope someone can help me out on this.

Thanks and regards,
Saba Taseer

Answer 1

I believe that the problem is that your typedef

typedef void (MyClass::Ptr2func)();

Is not defining a typedef for a pointer to member function, but for a member function type. The typedef for a member function pointer would be

typedef void (MyClass::* Ptr2func)();

Notice the explicit pointer involved here. The type

typedef void (MyClass::Ptr2func)();

is actually a typedef for the type of a member function inside of MyClass that takes no arguments and returns void.

As for your final question, the proper syntax for calling a pointer to a member function is (I believe)

(myCls.*var.ptr)()

Notice that you must parenthesize the expression (myClass.*var.ptr) before trying to call it as a function, since the code

myCls.*(var.ptr())

Means “dereference the pointer-to-member returned by var.ptr() relative to object myCls.

Hope this helps!