CODEWITHSUNDEEP
c++referenceoperator-overloadingreturn-typebrackets
A.P
A.P

Reputation: 3

Overloading bracket operator as member function

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? i.e: 

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)?

Upvotes: 0

Views: 406

Answers (1)

Tatsuyuki Ishi
Tatsuyuki Ishi

Reputation: 4031

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].

Upvotes: 2

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