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
– BrenBarn, Commented Jul 18, 2017 at 19:20

kichik · Accepted Answer · 2017-07-18 19:18:33Z

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, 2017 at 19:18

kichik

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2 Comments

Jared Goguen Over a year ago

might be worth adding that return dict(name=name, **kwargs) could be used to return all arguments

Hannan Over a year ago

Thanks, both answers helped me.

Cowabunghole · Accepted Answer · 2017-07-18 19:23:59Z

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, 2017 at 19:23

Cowabunghole

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