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*args and **opts oddities (?)

Michael Hudson mwh at python.net
Thu Nov 29 05:44:17 EST 2001 
Dinu Gherman <gherman at darwin.in-berlin.de> writes:

> Hi,
> 
> I'm trying to transform *args and **opts parameters and pretend
> they were passed in some transformed way. What I find is an un-
> expected phenomenon that makes me use indices in the code below
> (in function transform1) to get the behaviour I want. Is there 
> any good explanation for this?
> 
> Thanks,
> 
> Dinu
> 
> 
> # unexpected result: lines (1) and (2) are not equal
> 
> def transform0(*args, **opts):
>     print 't', args, opts
>     return args, opts 
> 
> def call0(*args, **opts):
>     print args, opts # (1)
>     args, opts = transform0(args, opts)

Think about what this does; it calls transform0 with two positional
arguments, the tuple `args', and the dictionary `opts'.  These then go
into the `*args' parameter of transform0, and the dictionary `opts' is
empty -- because there weren't any keyword arguments.

Did you want to write

    args, opts = transform0(*args, **opts)

?

Cheers,
M.

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