Question 6 in Object Oriented Python practices

Q&A
#1

Screen Link:

My Code: 

class SupermarketQueue:
    def __init__(self):
        self.queue=[]
    def add_to_back(self,element):
        self.queue.append(element)
    def remove_from_front(self):
        return self.queue.pop(0)
    def __str__(self):
        string=self.queue
        string.reverse()
        return str(string)
    def __len__(self):
        return len(self.queue)

queue=SupermarketQueue()
queue.add_to_back('Alice')
queue.add_to_back('Bob')
print(queue.__len__())
print(queue)
print(queue.remove_from_front())

What I expected to happen:

2
[‘Bob’, ‘Alice’]
Bob

The above output is correct, I don`t get why it says the below error message
What actually happened:
Method str returned the wrong value.

Replace this line with the output/error
#2

@shi.aaron

string=self.queue is the major problem. string points to the object self.queue. So when you do string.reverse(), it reverses self.queue.

This print(queue.remove_from_front()) prints Bob instead of Alice as a result.

I introduced copy and everything works fine.

import copy
class SupermarketQueue:
    def __init__(self):
        self.queue=[]
    def add_to_back(self,element):
        self.queue.append(element)
    def remove_from_front(self):
        return self.queue.pop(0)
    def __str__(self):
        string=copy.deepcopy(self.queue)
        string.reverse()
        return str(string)
    def __len__(self):
        return len(self.queue)

queue=SupermarketQueue()
queue.add_to_back('Alice')
queue.add_to_back('Bob')
print(queue.__len__())
print(queue)
print(queue.remove_from_front())

Without copy, both string and queue have the same address.

def __str__(self):
        string= self.queue
        string.reverse()
        print(id(string), id(self.queue))
        return str(string)

Ouput: 140505287815424 140505287815424
#3

I see where the issue is. Many thanks!

1 Like