python3: what exactly does `**` operator does

Question

def func(**kwargs):
    print(kwargs)

Are these two function calls identical?

d = {'id':1, 'name':'qwerty', 'dtype':3}
func1(**d)
func1(id=1, name='qwerty', dtype=3)

As far I understand, with a formal parameter in the form **arg, it receives a dictionary, while the formal parameter as *arg yields the tuple.

What is actually happening behind **kwargs?

Have a look at [What is the purpose and use of \*\*kwargs?](https://stackoverflow.com/questions/1769403/what-is-the-purpose-and-use-of-kwargs) — khelwood, Dec 13 '19 at 19:46
@khelwood, thanks for link. It explains everything I wanted to know about kwargs. — Mark, Dec 13 '19 at 19:57

score 0 · Answer 1 · answered Dec 13 '19 at 19:44

It does exactly as you said.

**kwargs allows you to pass keyworded variable length of arguments to a function.

You should use **kwargs if you want to handle named arguments in a function.

** takes a dictionary and "spreads" it to a function as keyword (named) params.

python3: what exactly does `**` operator does

1 Answers1