Issues using .join() and .pop() on one line

I just spent a lot of time trying to solve the following very low level coding challenge on Edabit.com:

This is a list of single characters with an unwanted character at the end:

["H", "e", "l", "l", "o", "!", "\0"]

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You could also just type "Hello!" when initializing a variable,
creating the string "Hello!"

Create a function that will return a string by combining the given
character list, not including the unwanted final character. Examples

cpp_txt(["H", "i", "!", "\0"]) ➞ "Hi!"

I couldn’t solve this trying the following one one line:

return "".join(lst.pop(-1))

or return "".join(lst.pop())

I get very strange results this way:

FAILED: '\x00' should equal 'Hi!'
ERROR: Traceback:
   in <module>
  File "./frameworks/python/cw-2.py", line 28, in assert_equals
    expect(actual == expected, message, allow_raise)
  File "./frameworks/python/cw-2.py", line 18, in expect
    raise AssertException(message)
cw-2.AssertException: '\x00' should equal 'Hi!'

I tried all kinds of things, like .strip(‘\x00’) but nothing worked. I used a for loop, I got results like ‘Hi!x00’.

Finally I gave up, and the solutions are either this:

def cpp_txt(lst):
return ''.join(lst[:-1])

that was probably what I should’ve thought of in the first place,
or this:

def cpp_txt(lst):
a = lst.pop(-1)
return "".join(lst)

Now, analyzing the second solution I guess I am using .pop() wrong, it seems like you need to create a separate variable to be able to ‘clean up’ the list, but this does not seem to be necessary in the following example I found in my browsing history on another website:

data_science_topics = ["Machine Learning", "SQL", "Pandas", "Algorithms", "Statistics", "Python 3"]
print(data_science_topics)

# Your code below: 
data_science_topics.pop()
print(data_science_topics)
data_science_topics.pop(3)
print(data_science_topics)

So, I hope someone would be so kind to explain me why the weird "\x00" appears in my results and how come I shouldn’t be using .join() and .pop() like this on a single line

Thank you for your time and patience.

>Solution :

When you do:

return "".join(lst.pop(-1))

It is equivalent of:

def cpp_txt(lst):
    a = lst.pop(-1)
    return "".join(a)  # NOTE: This is a ... not lst

In your working solution, you wanted to join the lst – i.e. return "".join(lst). But that is not what pop returns.

.pop() modifies the list provided to it – and returns just the last element.