Filling a shape with color in python turtle

Question

I'm trying to fill a shape with a color but when I run it, it does not show. Am I not supposed to use classes for this? I am not proficient with python-3 and still learning how to use classes

import turtle

t=turtle.Turtle()
t.speed(0)


class Star(turtle.Turtle):
    def __init__(self, x=0, y=0):
        turtle.Turtle.__init__(self)
        self.shape("")
        self.color("")
#Creates the star shape    
    def shape(self, x=0, y=0):
        self.fillcolor("red")
        for i in range(9):
        self.begin_fill()
        self.left(90)
        self.forward(90)
        self.right(130)
        self.forward(90)
        self.end_fill()
#I was hoping this would fill the inside        
    def octagon(self, x=0.0, y=0.0):
        turtle.Turtle.__init__(self)

    def octa(self): 
        self.fillcolor("green")
        self.begin_fill()
        self.left(25)
        for x in range(9):
            self.forward(77)
            self.right(40)

#doesn't run with out this
a=Star()

Answer 1

Issues with your program: you create and set the speed of a turtle that you don't actually use; turtle.py already has a shape() method so don't override it to mean something else, pick a new name; you don't want the begin_fill() and end_fill() inside the loop but rather surrounding the loop; you call your own shape() method with invalid arguments.

The following rework of your code addresses the above issues:

from turtle import Turtle, Screen

class Star(Turtle):
    def __init__(self, x=0, y=0):
        super().__init__(visible=False)
        self.speed('fastest')
        self.draw_star(x, y)

    def draw_star(self, x=0, y=0):
        """ Creates the star shape """

        self.penup()
        self.setposition(x, y)
        self.pendown()

        self.fillcolor("red")

        self.begin_fill()

        for _ in range(9):
            self.left(90)
            self.forward(90)
            self.right(130)
            self.forward(90)

        self.end_fill()

t = Star()

screen = Screen()
screen.exitonclick()