Re: [PyOpenGL-Users] Converting C++ code to python code

[Derakon <de...@gm...>]

On Sun, Apr 17, 2011 at 11:44 AM, Abhijeet Rastogi
<abh...@gm...> wrote:
>
> Original author has implemented figureSet as vector, I implemented it as
> python list while converting the code.  In the below code snippet, "in the
> if-else part", I am not able to make out, how will I convert the code in
> python?
>
> void drawSel() {
>   glColor3f(0, 0, 0);
>   int size = figureSet.size();
>   Figure *f = figureSet[selected];
>
>   /* getPoint() is a virtual method: every figure has at least
>       two points (is at least a line) */
>   int *pt1 = f->getPoint(1)->getCoords();
>   int *pt2 = f->getPoint(2)->getCoords();
>
>   if (Triangle *t = dynamic_cast<Triangle*>(f)) { // triangle: one more
> point
>      int *pt3 = t->getPoint(3)->getCoords();
>      //something happens here
>   }
>   else if (Quad *q = dynamic_cast<Quad*>(f)) { // quad: two more point
>      int *pt3 = q->getPoint(3)->getCoords();
>      int *pt4 = q->getPoint(4)->getCoords();
>      //Something happens here
>   }
> }

In this code we have a guarantee that there are at least 2 points to
draw, but if the figure is a triangle, then there's only one more,
while if it's a quad then there's two more. These need different
drawing logic, hence the if/else split. The conditionals are basically
saying "How many points are in this polygon?" and then drawing
differently depending on the result. In other words, dynamic_cast is
being used to check if the figure variable is representing a Triangle
or a Quad.

The direct way to do this in Python would be something along the lines of

if type(figure) == Triangle:
    draw as if it's a triangle
elif type(figure) == Quad:
    draw as if it's a quad

Assuming you have Triangle and Quad classes declared, of course.

The more elegant way would be to have Triangle and Quad have their own
draw functions; then you just do figure.draw() and the appropriate
function is automatically invoked.

-Chris