forwarding *arg parameter

[Steven D'Aprano]

On Sun, 05 Nov 2006 19:35:58 +0000, Tuomas wrote:

> Thanks. My solution became:
> 
>  >>> def flattern(arg):
> ...     result = []
> ...     for item in arg:
> ...         if isinstance(item, (list, tuple)):
> ...             result.extend(flattern(item))
> ...         else:
> ...             result.append(item)
> ...     return tuple(result)
> ...
>  >>> def g(*arg):
> ...     arg = flattern(arg)
> ...     return arg
> ...
>  >>> def f(*arg):
> ...     return g(arg)
> ...
>  >>> f('foo', 'bar')
> ('foo', 'bar')


That's the most complicated do-nothing function I've ever seen. Here is a
shorter version:

def shortf(*args):
    return args


>>> f('foo', 'bar')
('foo', 'bar')
>>> shortf('foo', 'bar')
('foo', 'bar')

>>> f(1,2,3,4)
(1, 2, 3, 4)
>>> shortf(1,2,3,4)
(1, 2, 3, 4)

>>> f({}, None, 1, -1.2, "hello world")
({}, None, 1, -1.2, 'hello world')
>>> shortf({}, None, 1, -1.2, "hello world")
({}, None, 1, -1.2, 'hello world')

Actually, they aren't *quite* identical: your function rips lists apart,
which is probably not a good idea.

>>> f("foo", [1,2,3], None) # three arguments turns into five
('foo', 1, 2, 3, None)
>>> shortf("foo", [1,2,3], None) # three arguments stays three
('foo', [1, 2, 3], None)



I still don't understand why you are doing this. Can we have an example of
why you think you need to do this?



-- 
Steven.