Passing one list instead of two in python

Question

Here I have written a function which takes two lists as argument. But when I called this function passing one list as argument, it works well! Why is this working? Here name_function has two arguments but I passed only one list as argument.

def name_function(names=list(),_list=list()):
    for name in names:
        _list.append(name)
    return _list
print(name_function(['mike','smith','bob']))

If you pass only a single list, then the second list is automatically set to an empty list per: `_list=list()`. — JonSG, Jun 15 '21 at 16:32
`names=list()` means that you are creating empty list and passing it to name parameter as default value. Same thing done with _list. — Rima, Jun 15 '21 at 16:40
Does this answer your question? ["Least Astonishment" and the Mutable Default Argument](https://stackoverflow.com/questions/1132941/least-astonishment-and-the-mutable-default-argument) — L3viathan, Jun 15 '21 at 16:52
@L3viathan I think OP is asking why default arguments work in the first place. But with mutable defaults, I'm sure that question will be relevant soon enough. ;) — Axe319, Jun 15 '21 at 17:48
@dibery yes I have now got this. When two lists are arguments, after passing one list , the later would be default one. — roktim, Jun 15 '21 at 18:42

score 0 · Answer 1 · answered Jun 17 '21 at 05:17

Here in the function definition you have initialized both the arguments with the empty list i.e. you are using two default arguments in the function definition.
For the above reason the function call works if you provide provide both the arguments or any one of the arguments or no argument at all.
To learn more about default arguments in python you can refer this link or this link.

Passing one list instead of two in python

1 Answers1