Python optional list argument pulling in reference for another list

Question

I ran into a very strange bug where an argument being initialized to an empty list, is grabbing a different global list. Can someone explain why this is happening?

def append_letters(s,letters,history_local=list()): 
    s += letters
    # Record delete action for undo
    history_local.append((2,len(letters)))
    return s, history_local

history = [(2, 2), (1, 'cdefg')]
s = 'ab'
hist = history.pop()
s = append_letters(s,hist[1])
print(history)
# Expect - [(2,2)] - only the un-popped part of history
# Result - [(2,2),(2,5)] - history is modified by reference even though it is neither returned nor passed to append_letters

What version of python are you using? When I run this same code under `python v3.9.6` I get the result: `[(2,2)]` — dalekman1234, Aug 02 '22 at 17:57
It's not a global list; it's the single list `append_letters` uses as the default value for `append_letters`, and your function mutates that list. — chepner, Aug 02 '22 at 18:57
`append_letters.__defaults__[0]` produces a reference to that list. — chepner, Aug 02 '22 at 18:58

score 0 · Answer 1 · answered Aug 02 '22 at 18:52

0

The issue was being caused by append_letters returning two arguments - s, and history which were being interpreted incorrectly at a higher level.

answered Aug 02 '22 at 18:52

illan

163
1
13

Python optional list argument pulling in reference for another list

1 Answers1