Overloading bracket operator as member function

Question

I'm making a simple class for Vector3D type objects. The following code will compile and run perfectly.

class Vector3D {
    float x, y, z;
public:
    Vector3D() = default;
    Vector3D(float a, float b, float c) : x(a), y(b), z(c) {};
    float& operator [](int i) {
        return (&x)[i];
    }
    const float& operator [](int i) const {
        return (&x)[i];
    }
}

int main(int argc, char ** argv){
    Vector3D myVec(1,2,3);
    printf("Value of y: %d\n", myVec[1]);
}

However when I remove the address-of operator (&) I get an error and the code will not work. Why is the (&) necessary? ie:

return (x)[i]; // will not compile "expression must have pointer-to-object type"
return (&x)[i]; // all good here

Also I'm having trouble understanding how this even works. How is it that the function can return the ith float, do member variables get stored in a contiguous way in memory (like arrays)?

Answer 1

The way you're doing this is so tricky, and this is undefined behavior .

There is no guarantee for struct member layout, but most time the members are placed on memory as:

x---y---z--- (4 bytes each)
x[0]
    x[1]
        x[2]

So this is why your code is working (remember this is not a defined behavior).

Your code doesn't do bounds checking anyway, so consider:

1. Converting it into a switch.
2. Make your member into an array like float x[3] .