Python functions with default arguments

Question

I'm learning Python and playing around with function arguments. Just created a test function with following code:

def packer(name = 'Alpha', **kwargs):
    return(kwargs)

And if I call this function as:

dummy_packer = packer(name = 'Bravo', age = 65, beard = False)

The result is:

{'age': 65, 'beard': False}

The variable dummy_packer will not have name value at all in the result. I understand it ignored it because I have defined already at the stage of creating the function. But then why it didn't give me the default value also? Where the name argument is stored?

Thanks

The `name` argument is stored in the `name` variable. `**kwargs` doesn't mean "all arguments passed by keyword", it means "all arguments passed by keyword that aren't named in the function signature". — BrenBarn, Jul 18 '17 at 19:20
See https://stackoverflow.com/questions/3394835/args-and-kwargs — akshaynagpal, Jul 18 '17 at 19:21
Possible duplicate of [\*args and \*\*kwargs?](https://stackoverflow.com/questions/3394835/args-and-kwargs) — akshaynagpal, Jul 18 '17 at 19:21

score 2 · Accepted Answer · answered Jul 18 '17 at 19:18

2

name is available to you in the function context.

def packer(name = 'Alpha', **kwargs):
    print('name is %s' % (name,))
    return(kwargs)

answered Jul 18 '17 at 19:18

kichik

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might be worth adding that `return dict(name=name, **kwargs)` could be used to return all arguments – Jared Goguen Jul 18 '17 at 19:26
Thanks, both answers helped me. – Hannan Jul 18 '17 at 19:39

score 0 · Answer 2 · answered Jul 18 '17 at 19:23

0

kwargs refers to variable number of keyword arguments. In this case, because name is a defined variable, it is not included in kwargs.

answered Jul 18 '17 at 19:23

Cowabunghole

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Python functions with default arguments

2 Answers2