C++ call method on object with the object itself as parameter - c++

Question

In python you can, for example, call array.sort() and it will sort the array it is called on. However, I now have the following code snippet
void drawClickableRectangle(ClickableRectangle recto){
ofSetHexColor(0xffffff); // just some syntax from the library I'm using
ofFill();
ofDrawRectangle(recto.xpos, recto.ypos, recto.width, recto.height);
}
and then call this method here:
ClickableRectangle recto(1,1,100,100);
recto.drawClickableRectangle(recto);
This is the full class:
class ClickableRectangle
{
// Access specifier
public:
// Data Members
int xpos, ypos, width, height;
ClickableRectangle(int x1, int y1, int width1, int height1){
xpos = x1;
ypos = y1;
width = width1;
height = height1;
};
// Member Functions()
int getxpos()
{
return xpos;
}
int getypos(){
return ypos;
}
int getwidth(){
return width;
}
void drawClickableRectangle(ClickableRectangle recto){
ofSetHexColor(0xffffff);
ofFill();
ofRect(recto.xpos,recto.ypos, recto.width, recto.height);
//ofDrawRectangle(recto.xpos, recto.ypos, recto.width, recto.height);
}
Is there a way to make the function call "reflexive"? So I can just call it like:
recto.drawClickableRectange();
I'm relatively new to C++, but not to programming in general. Thanks!

Answer 1

You can do like this in C++:
class ClickableRectangle {
public int xpos;
public int ypos;
public int width;
public int height;
void drawClickableRectangle(){
ofSetHexColor(0xffffff); // just some syntax from the library I'm using
ofFill();
ofDrawRectangle(xpos, ypos, width, height);
}
}
Then in your main function, call it like this:
int main(){
ClickableRectangle recto;
recto.xpos = 1;
recto.ypos = 1;
recto.width = 100;
recto.height = 100;
recto.drawClickableRectange();
return 0;
}

Answer 2

Not in the same way as python, no.
In python, you can
def unattached(fake_self):
return fake_self.x
class Thing:
def __init__(self):
self.x = 42
Thing.method = unattached
thing = Thing()
print (thing.method())
print (unattached(thing))
Because there's no difference between a free function with an explicit first paramter and a instance method with an implicit first parameter.
In C++, you can't change a class at runtime, and a member function has a different type to a free function.
struct Thing {
int x = 42;
int method() const { return this->x; }
}
int unattached(const Thing * thing) { return thing->x; }
The type of unattached is int (*)(const Thing *), whereas method is a int (const Thing::*)(). These are different types, you can't switch one for the other. You can however construct a std::function<int(const Thing *)> from either of them, but you can only use that with the free function syntax func(thing), as it isn't a member of Thing