[Python-Dev] chained assignment weirdity

Terry Reedy tjreedy at udel.edu
Tue Nov 6 10:24:37 CET 2012

On 11/6/2012 1:18 AM, Chris Withers wrote:
> Hi All,
> I bumped into this using Michael Foord's Mock library.
> It feels like a bug to me, but thought I'd ask here before logging one
> in the tracker in case people know that we won't be able to fix it:
> On 05/11/2012 13:43, Michael Foord wrote:
>>>>>>> class Foo(object):
>>>> ...  def __setattr__(s, k, v):
>>>> ...   print k
>>>> ...
>>>>>>> f = Foo()
>>>>>>> f.g = f.x = 3
>>>> g
>>>> x
>>> Hmm, that's definitely counter-intuitive. Worth raising on python-dev?
>> Well, I guess "it's always been that way", so changing it would be
>> backwards incompatible. Feel free to raise it though. :-)

You are expecting a chained assignment a = b = c to be parsed as
a = (b = c), as in C and as it should be if assignment is an expression. 
But in Python it is not. The right association would be meaningless 
since (b = c) has no value to be assigned to a.

Python actually parses the chained assignment as (a =) (b =) c.
The relevant grammar production is "assignment_stmt ::=
(target_list "=")+ (expression_list | yield_expression)".
The explanation after the grammar begins with this sentence.

"An assignment statement evaluates the expression list (remember that 
this can be a single expression or a comma-separated list, the latter 
yielding a tuple) and assigns the single resulting object to each of the 
target lists, from left to right."

In other words, in my example, c is assigned to a then b. Please do not 
report documented behavior on the tracker as a bug (and don't encourage 
people to). If you think the above is not clear enough, you *can* 
suggest improvement. Perhaps add an example and explanation such as

a = b, (c,d), *e, f = 1, (2, 3), 4, 5, 6

"The tuple on the right is first assigned to a and then unpacked to b, 
c, d, e, and f, giving them the values 1, 2, 3, [4, 5], and 6."

Terry Jan Reedy

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