There are a few things to know when you write operators, which are not as often used with other functions.
The assign operators, for example, will return *this because you change the value of the vector:
class v {
public:
double x_, y_;
v& operator += (const v& rhs)
{
_x += rhs._x;
_y += rhs._y;
return *this;
}
};
Another interesting one, the pre ++ and post ++ are different only because of an unused parameter:
class v {
public:
double x_, y_;
v& operator ++ (); // ++v
v& operator ++ (int); // v++
};
The "equal" (assignment) is another one that is tricky when you use pointers. For a vector, it generally won't be a problem, but if you define a vector V and assign it to itself, you have to be careful:
class v {
public:
double x_, y_;
v& operator = (const v& rhs)
{
if(this != &rhs)
{
x_ = rhs.x_;
y_ = rhs.y_;
}
return *this;
}
};
In your case, the if() will most certainly not be useful, but think about doing something like this:
delete p_;
p_ = new foo;
p_->x_ = rhs.p_->x_;
If &rhs == this, then the delete p_ deleted the rhs pointer! That means accessing it on the 3rd line is a bug.
The rest should be easy enough to work with. The compare operators return bool and are const:
class v {
public:
double x_, y_;
bool operator == (const v& rhs) const
{
return x_ == rhs.x_ && y_ == rhs.y_;
}
};
Although, since C++20, you are expected to only declare the three way comparison operator <=> which allows the compiler to implement all the other comparison operators for you. This one returns a negative number (smaller: a < b), 0 (equal: a == b), or a positive number (larger: a > b).
I'm not sure what makes a vector bigger or smaller, I used the length from (0, 0) in this example:
class v {
public:
double x_, y_;
int operator <=> (const v& rhs) const
{
if(x_ == rhs.x_ && y_ == rhs.y_)
{
return 0;
}
return length() > rhs.length() ? 1 : -1;
}
};
Except for the [] operator. There are two versions of that one:
class v {
public:
// I would imagine you'd use an array but as a simple example...
double x_, y_;
double operator [] (int idx) const
{
return idx == 0 ? x_ : y_;
}
v_ref operator [] (int idx)
{
v_ref v(this, idx);
return v;
}
};
As you can see, the non-constant version of the [] operator returns a reference. This is necessary so you can write something like:
r[3] = 7.3;
r[3] returns that reference, then the assignment of the reference is called with 7.3 as the parameter. (Note that we should probably throw an error if you use 3 as the index when you only have 2 values: 0 and 1--this is not shown here)
class v_ref
{
public:
v *p_;
int i_;
v_ref(v *p, int i)
: p_(p), i_(i)
{
}
operator = (double q)
{
// again, I suppose you'd use an array instead!
if(i_ == 0)
{
p_->x_ = q;
}
else
{
p_->y_ = q;
}
}
};
Assuming you want some security, the vector pointer could make use of a reference counter so you know whether a main vector object gets deleted before all of its reference objects...
Another note: I would imagine that your constructor will allocate an array of double (or use an std::vector<double> type...) If you use new, remember to delete in the destructor and that's when the if() in the assignment operator is very important.
