Passing all arguments of a function to another function
arguments, function, python
Solution
Explicit is better than implicit but if you really don't want to type a few characters:
def func1(a=1, b=2, c=3):
func2(**locals())
`locals()` are all local variables, so you can't set any extra vars before calling `func2` or they will get passed too.
Problem
I want to pass all the arguments passed to a function(`func1`) as arguments to another function(`func2`) inside `func1` This can be done with `*args, *kwargs` in the called `func1` and passing them down to `func2`, but is there another way? Originally ``` def func1(*args, **kwargs): func2(*args, **kwargs) ``` but if my func1 signature is ``` def func1(a=1, b=2, c=3): ``` how do I send them all to func2, without using ``` def func1(a=1, b=2, c=3): func2(a, b, c) ``` Is there a way as in javascript `callee.arguments`?