**kwargs

[sp00fD]

In article <1263886656-955174 at hypernet.com>,
  gmcm at hypernet.com wrote:
> sp00fD writes:
>
> > my.foo(foo="something", bar="something else", this_dict)
>
> Not quite. To call foo with the named args wrapped up as a
> dictionary, you need to use apply:
>   apply(foo, ("something", "something else"), this_dict)

Using the above apply(), how does foo know that foo(should have named
that something else) == "something" and bar == "something else", they
appear to be passed as normal args.

Also, how is this dictionary represented after the foo method receives
it?  ie. this_dict["wow"] = 1
apply(foo, (....), this_dict)

foo then sees kwargs["wow"] == 1?

> or
>   this_dict['foo'] = "something"
>   this_dict['bar'] = "something else"
>   apply(foo, (), this_dict)
>
> Note that foo cannot tell which of the above ways was used,
> (ie, "foo" and "bar" are not in kwargs when foo examines it).
>
> BTW, while foo-abuse is common, I've never seen it carried to
> such extremes.

I do like foo! ;)
Thanks
>
> - Gordon
>
>


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