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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 ...
#1: Initial revision
by
Karl Knechtel
·
2023-09-20T03:45:39Z (about 1 year ago)
Understanding mutable default arguments in Python
Consider this code example:
```
def example(param=[]):
param.append('value')
print(param)
```
<details><summary>
When `example` is repeatedly called with an existing list, it repeatedly appends to the list, as one might expect:</summary>
```
>>> my_list = []
>>> example(my_list)
['value']
>>> example(my_list)
['value', 'value']
>>> example(my_list)
['value', 'value', 'value']
>>> my_list
['value', 'value', 'value']
```
</details><details><summary>If it's called with an empty list each time, it seems that each empty list is considered separately, which also makes sense:</summary>
```
>>> example([])
['value']
>>> example([])
['value']
>>> example([])
['value']
```
</details><details><summary>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:</summary>
```
>>> example()
['value']
>>> example()
['value', 'value']
>>> example()
['value', 'value', 'value']
```
</details>
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?