**args in function raises TypeError "takes 0 positional arguments" [python]

Question

def f(**args):
    print(args)

This gives TypeError exception. Though when I pass no args, like f(). It just prints nothing.

What could be the reason and how to pass args? I tried with f( [4,5]), but still same result

Answer 1

f only has a dictionary argument **args (more frequently written as **kwargs), which means that only keyword arguments are allowed (passing nothing at all is of course also allowed).

To enable positional arguments, add *args:

def f(*args,**kwargs)

or pass your parameter with any keyword:

f(foo=[4,5])

and kwargs["foo"] is [4,5] in f

Answer 2

** handles keyword arguments. So your function indeed doesn't take any positional arguments, only keyword arguments (or nothing).

Positional arguments are aggregated with *args, conventionally keyword arguments with **kwargs.