Use C structs only and stay OOPy?

Question

Say you have:

struct c_struct {
   int value;
   /* other stuff */
   void (* dump)();
};

and you'd like to, at some point:

c_struct_obj->dump();

I assume there's no way you could instantiate a c_struct object such that its particular "dump" function knows its particular "value" the way C++ methods know member variables (via the implicit "this" I suppose)? I guess I know the answer already ("no"). If so, are there other ways of using C structs in an OOPy way?

Answer 1

Sure, you just have to pass this by yourself:

struct c_struct {
    int value;
    /* other stuff */
    void (* dump)(struct c_struct *this);
};

And then call it with:

c_struct_obj->dump(c_struct_obj);

Answer 2

You can use structs in this way but its a pain. All c++ member function get a *this pointer passed to them you can do the same but your dump functions will need to take the structure that its contained in as a parameter.

Answer 3

If you are using a C++ compiler, the only difference between a struct and a class is the default visibility of member variables (classes being private, structs being public). The this pointer is available within member functions.

struct test
{
    int x;

    void inc();
};

void test::inc()
{
    x++;
}

int main(void)
{
    test a;
    a.x = 1;
    a.inc();
    int b = a.x;

    return 0;
}

b == 2 here.

Answer 4

You can do this and can even do inheritance. But the interfacing with it is a total mess and you get nowhere near, for example, the resource safety involved with C++'s deterministic automatic cleanup. Ultimately, you CAN do OOP in C, but it's just not worth the hassle compared to just using C++, as well as all the other features that C++ offers.