[Python-ideas] Add .= as a method return value assignment operator

[Brice Parent]

Le 27/09/2018 à 12:48, Ken Hilton a écrit :
> Hi Jasper,
> This seems like a great idea! It looks so much cleaner, too.
>
> Would there be a dunder method handling this? Or since it's explicitly 
> just a syntax for "obj = obj.method()" is that not necessary?
> My only qualm is that this might get PHP users confused; that's really 
> not an issue, though, since Python is not PHP.
>
> Anyway, I fully support this idea.
>
What would the following evaluate to?
a .= b + c(d)

1: a = a.b + a.c(a.d)  # everything is prepended an "a."
it means we dn't have access to any external elements, making the 
functionality only useful in a few cases

2: a = a.b + a.c(d)  # every first level element (if that means 
something) is prepended an "a."
We still lose some of the ability to access anything outside of `a`, but 
a bit less than in #1. The effort to understand the line as grown a bit, 
though.

3: a = a.(b + c(d))  # everything is evaluated, and an "a." is prepended 
to that result
(the same way `a *= 2 + 3` is equivalent to `a *= 5`)
I believe in most cases, this wouldn't mean anything to evaluate `b + 
c(d)` on their own, and expect a return that can be used as an attribute 
of `a`.

4: a = a.b + c(d)  # "a." is prepended to the first element after the `=`
It is probably quite easy to read and understand, but it removes the 
transitivity of the operators we have on the right, and is a bit limiting.

5: SyntaxError: Can only use the [whatever the name] augmented operator 
with a single expression
Why not, it's a bit limiting, but is clear enough to me.

Maybe, a simpler thing to do for this problem would be to make something 
like this:
a = .b(5) + c(.d) + 3
being the equivalent of
a = a.b(5) + c(a.d) + 3

I don't see any ambiguity anymore, it shortens the code a lot, and I 
guess it wouldn't be hard for the compiler to recompose the line as a 
first parsing step, and create the same AST with both syntaxes.
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