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c++multiple-inheritancereinterpret-cast

Multiple inheritance and the this pointer


Suppose I have this struct:

struct vector_data
{
    double x, y;

    double& operator[](size_t index)
    {
        return * (static_cast<double*>(static_cast<void*>(this)) + index);
    }
};

The operator[] should work as expected, because vector_data is a POD type. The expected behaviour is that vector_data[0] returns x, and vector_data[1] returns y.

Now suppose I have a second struct:

struct more_data
{
    double evil_data;

    // There could be more here, data or functions
};

And derive from both like this:

struct composed : public more_data, public vector_data
{
};

Will this destory the expected behaviour of operator[]? In other words, will the this-pointer of vector_data in the derived struct still point to the vector_data part of the struct, or will it point to the beginning of the derived struct?

If it does destroy operator[], then how can I resolve this problem? I can inherit from vector_data first, but suppose composed contains virtual functions. I know most compilers put the vtable at the end, but this is not guaranteed. What would be the best approach?


Solution

  • Leaving aside the issues of your incorrect pointer arithmetics (the possibility of padding between x and y invalidates your assumption), here is a quick illustration of what's going on with this pointer when you use multiple inheritance:

    #include <iostream>
    using namespace std;
    
    struct a {
        int aa;
        void showA() {
            cerr << this << endl;
        }
    };
    struct b {
        int bb;
        void showB() {
            cerr << this << endl;
        }
    };
    struct c : public a, b {
        int cc;
        void showC() {
            cerr << this << endl;
        }
    };
    int main() {
        c x;
        x.showA();
        x.showB();
        x.showC();
    }
    

    showA and showB print different numbers; showC prints the same number as showA, because a is listed first in the list of bases. If you switch a and b there, then showC and showB would be the same. The "magic" is in the C++ compiler: it is smart enough to give each member function a correct this pointer.