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Groups > comp.lang.python > #87796

__iadd__ with 2 args?

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I can write a member 

F.__iadd__ (self, *args)

and then call it with 2 args:

f = F()
f.__iadd__ (a, b)

And then args is:
(a, b)

But it seems impossible to spell this as

f += a, b

That, instead, results in 

args = ((a, b),)

So should I just abandon the idea that += could be used this way?

-- 
Those who fail to understand recursion are doomed to repeat it

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Thread

__iadd__ with 2 args? Neal Becker <ndbecker2@gmail.com> - 2015-03-20 17:25 -0400
  Re: __iadd__ with 2 args? Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2015-03-21 16:37 +1100

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