How to get all named args in a dict?

[kj]

In <mailman.113.1242254593.8015.python-list at python.org> Terry Reedy <tjreedy at udel.edu> writes:

>kj wrote:
>> 
>> Suppose I have the following:
>> 
>> def foo(x=None, y=None, z=None):
>>     d = {"x": x, "y": y, "z": z}
>>     return bar(d)
>> 
>> I.e. foo takes a whole bunch of named arguments and ends up calling
>> a function bar that takes a single dictionary as argument, and this
>> dictionary has the same keys as in foo's signature, so to speak.
>> 
>> Is there some builtin variable that would be the same as the variable
>> d, and would thus obviate the need to explicitly bind d?

>Use the built-in function locals()
> >>> def f(a,b):
>	x=locals()
>	print(x)

> >>> f(1,2)
>{'a': 1, 'b': 2}

That's *exactly* what I was looking for.  Thanks!

kynn


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