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
--
https://mail.python.org/mailman/listinfo/python-list
