Mutability of the **kwargs argument in Python

Question

Consider a case where I change the kwargs dict inside a method:

def print_arg(**kwargs):
    print kwargs.pop('key')

If I call the method pop_arg with a dictionary like this:

mydict = {'key':'value'}
print_arg(**mydict)

will mydict be changed by this call?

I am also interested in a more detailed explanation of the underlying method calling mechanism that lets mydict change or not.

Answer 1

No, mydict won't be changed. kwargs get unpacked into a new dictionary.

Consider the case where you have:

def print_arg(key=1, **kwargs):
    print(key)
    print(kwargs)

print_arg(**{'key':2,'foo':3,'bar':4})

In this case, it's obvious that kwargs is a different dict than you pass in because when it gets unpacked, it's missing the 'key' key.

Answer 2

@mgilson's answer is correct. But you should also be aware about the shallow copy.

def print_arg(**kwargs):
    print kwargs.pop('key')
    kwargs['list'].pop()  # will affect the original
    kwargs['dict'].pop('a') #will affect the original

mydict = {'key':'value', 'list':[2,3,4,5] ,'dict': { 'a':1,'b':2 }}
print_arg(**mydict)
print (mydict)  # Output : {'dict': {'b': 2}, 'list': [2, 3, 4], 'key': 'value'}

http://codepad.org/diz38tWF