Is it always safe to modify the `**kwargs` dictionary?

Question

Using the Python function syntax def f(**kwargs), in the function a keyword argument dictionary kwargs is created, and dictionaries are mutable, so the

Answer 1

It is always safe. As the spec says

If the form “**identifier” is present, it is initialized to a new ordered mapping receiving any excess keyword arguments, defaulting to a new empty mapping of the same type.

Emphasis added.

You are always guaranteed to get a new mapping-object inside the callable. See this example

def f(**kwargs):
    print((id(kwargs), kwargs))

kwargs = {'foo': 'bar'}
print(id(kwargs))
# 140185018984344
f(**kwargs)
# (140185036822856, {'foo': 'bar'})

So, although f may modify an object that is passed via **, it can't modify the caller's **-object itself.

---

Update: Since you asked about corner cases, here is a special hell for you that does in fact modify the caller's kwargs:

def f(**kwargs):
    kwargs['recursive!']['recursive!'] = 'Look ma, recursive!'

kwargs = {}
kwargs['recursive!'] = kwargs
f(**kwargs)
assert kwargs['recursive!'] == 'Look ma, recursive!'

This you probably won't see in the wild, though.