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Python Object-Oriented Python Dice Roller RPG Roller

I have been stuck here for almost an hour now.The feed back I get is 'Hand' doesn't have a roll method.What should I do?

Please help.

dice.py
import random


class Die:
    def __init__(self, sides=2):
        if sides < 2:
            raise ValueError("Can't have fewer than two sides")
        self.sides = sides
        self.value = random.randint(1, sides)

    def __int__(self):
        return self.value

    def __add__(self, other):
        return int(self) + other

    def __radd__(self, other):
        return self + other
class D20(Die):
    def __init__(self):
        super().__init__(sides=20)

from dice import D20

class Hand(list):
    def __init__(self, size=0, die_class=None, *args, **kwargs):
        self.size = size
        super().__init__()
        for _ in range(size):
            self.append(die_class())

    @classmethod
    def roll(cls, size, die_class=D20):
        return cls(size=size, die_class=die_class)

    @property
    def total(self):
        return sum(self)
hands.py
class Hand(list):
    @property
    def total(self):
        return sum(self)

1 Answer

boi
boi
14,242 Points

You've overcomplicated things, first of all, don't try to modify the code given to you by the teacher, here you modified the Hand class, which is not a good sign. Try implementing these instructions 👇

1) Don't modify the given Hand code, leave it as it is. 2) Read the instructions carefully, it says that there should be 2 parameters in the roll method, cls and the number of dice, so cls and NOD.

class Hand(list):
    @property
    def total(self):
        return sum(self)

    @classmethod
    def roll(cls, NOD): 👈 #Two parameters required as hinted in the challenge notes by the teacher.

Next the notes say "Inside the roll classmethod create a new instance of the Hand class and assign it to a variable", I think you are well aware of creating instances of any class, just follow along with the notes one sentence at a time, think a little, if you are having trouble msg me again here. I'll help you an infinite amount of times no problem for me.