User-defined augmented assignment

Tom Anderson twic at
Sun Oct 2 01:58:05 CEST 2005

On Thu, 29 Sep 2005, Pierre Barbier de Reuille wrote:

> a discussion began on python-dev about this. It began by a bug report, 
> but is shifted and it now belongs to this discussion group.
> The problem I find with augmented assignment is it's too complex, it's
> badly explained, it's error-prone. And most of all, I don't see any
> use-case for it !
> The most common error is to consider that :
> a += b <==> a.__iadd__(b)
> when the truth is :
> a += b <==> a = a.__iadd__(b)
> which can be very confusing, as the two "a" are not necessarily the
> same.

Indeed. I certainly didn't realise that was how it worked.

> So, what I would suggest is to drop the user-defined augmented 
> assignment and to ensure this equivalence :
> a X= b <=> a = a X b
> with 'X' begin one of the operators.

That seems quite an odd move. Your proposal would lead to even more 
surprising behaviour; consider this:

a = [1, 2, 3]
b = a
a += [4, 5, 6]
print b

At present, this prints [1, 2, 3, 4, 5, 6]; if we were to follow your 
suggestion, it would be [1, 2, 3].

So, -1, i'm afraid.

I think the right solution here is staring us in the face: if everyone 
seems to think that:

a += b <==> a.__iadd__(b)

Then why not make it so that:

a += b <==> a.__iadd__(b)

Principle of Least Surprise and all that.

Since not everything that should support += is mutable (integers, for 
example), how about saying that if the recipient of a += doesn't have an 
__iadd__ method, execution falls back to:

a = a + b

I say 'a + b', because that means we invoke __add__ and __radd__ 
appropriately; indeed, the __add__ vs __radd__ thing is a precedent for 
this sort of fallback.

Doesn't that leave everyone happy?


I'm not quite sure how that works but I like it ...

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