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Re: Multiple Assignment a = b = c

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  Re: Multiple Assignment a = b = c Terry Reedy <tjreedy@udel.edu> - 2016-02-16 14:31 -0500

#103026 — Re: Multiple Assignment a = b = c

On 2/16/2016 7:46 AM, srinivas devaki wrote:
> Hi,
>
> a = b = c
>
> as an assignment doesn't return anything, i ruled out a = b = c as
> chained assignment, like a = (b = c)
> SO i thought, a = b = c is resolved as
> a, b = [c, c]

https://docs.python.org/3/reference/simple_stmts.html#assignment-statements
"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."

a = b = c is the same as tem = c; a = tem; b = tem.

This does not work if tem is an iterator.

 >>> def g():
	yield 1
	yield 2

 >>> a,b = c,d = g()
Traceback (most recent call last):
   File "<pyshell#4>", line 1, in <module>
     a,b = c,d = g()
ValueError: not enough values to unpack (expected 2, got 0)
 >>> a, b
(1, 2)
 >>> c,d
Traceback (most recent call last):
   File "<pyshell#6>", line 1, in <module>
     c,d
NameError: name 'c' is not defined

> at-least i fixed in my mind that every assignment like operation in
> python is done with references and then the references are binded to
> the named variables.
> like globals()['a'] = result()
>
> but today i learned that this is not the case with great pain(7 hours
> of debugging.)
>
> class Mytest(object):
>      def __init__(self, a):
>          self.a = a
>      def __getitem__(self, k):
>          print('__getitem__', k)
>          return self.a[k]
>      def __setitem__(self, k, v):
>          print('__setitem__', k, v)
>          self.a[k] = v
>
> roots = Mytest([0, 1, 2, 3, 4, 5, 6, 7, 8])
> a = 4
> roots[4] = 6
> a = roots[a] = roots[roots[a]]

tem = roots[roots[a]]
a = tem
roots[a] = tem

> the above program's output is
> __setitem__ 4 6
> __getitem__ 4
> __getitem__ 6
> __setitem__ 6 6

-- 
Terry Jan Reedy

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