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Comments on Understanding mutable default arguments in Python
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Understanding mutable default arguments in Python
Consider this code example:
def example(param=[]):
param.append('value')
print(param)
When example is repeatedly called with an existing list, it repeatedly appends to the list, as one might expect:
example is repeatedly called with an existing list, it repeatedly appends to the list, as one might expect:>>> my_list = []
>>> example(my_list)
['value']
>>> example(my_list)
['value', 'value']
>>> example(my_list)
['value', 'value', 'value']
>>> my_list
['value', 'value', 'value']
If it's called with an empty list each time, it seems that each empty list is considered separately, which also makes sense:
>>> example([])
['value']
>>> example([])
['value']
>>> example([])
['value']
However, if called without passing an argument - using the default - it seems to "accumulate" the appended values as if the same list were being reused:
>>> example()
['value']
>>> example()
['value', 'value']
>>> example()
['value', 'value', 'value']
Some IDEs will warn that param is a "mutable default argument" and suggest changes.
Exactly what does "mutable default argument" mean, and what are the consequences of defining param this way? How is this functionality implemented, and what is the design decision behind it? Is it ever useful? How can I avoid bugs caused this way - in particular, how can I make the function work with a new list each time?
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