How Python's a, b = b, a works?

Question

i know that my will seem trivial,but i cant figure out why isn't 'stop' equal to zero, since 'start' value is already overwritten.But when i simulate the same scenario out of the function it does indeed overwrite it.Am i missing something here?

def interval(start, stop =None, step = 1 ):
    'Imitates range() for step >0 '
    if stop is None:
        start, stop = 0, start #since start is already 0, why isn't stop as well?
        print start,stop , step
    result = []
    i = start
    while i< stop:
        result.append(i)
        i+=step 
    return result

thefourtheye · Accepted Answer · 2014-12-11T04:22:22.903

This expression

start, stop = 0, start

will NOT be evaluated like this

start = 0
stop = start

but will be evaluated by pushing both the values to the stack and the top two values will be rotated (so that the values will be swapped) and the values are assigned back to start and stop. Lets consider,

a = 1
a, b = 0, a

If we look at the disassembled code

import dis
dis.dis(compile("a = 1; a, b = 0, a", "<string>", "exec"))

it would look like

  1           0 LOAD_CONST               0 (1)
              3 STORE_NAME               0 (a)
              6 LOAD_CONST               1 (0)
              9 LOAD_NAME                0 (a)
             12 ROT_TWO
             13 STORE_NAME               0 (a)
             16 STORE_NAME               1 (b)
             19 LOAD_CONST               2 (None)
             22 RETURN_VALUE

The LOAD_* operations will load the content in to the stack. If you look at

              6 LOAD_CONST               1 (0)
              9 LOAD_NAME                0 (a)

the values, 0 and the value in a are pushed to the stack and then ROT_TWO, rotates the values in the top two positions of the stack. And finally

             13 STORE_NAME               0 (a)
             16 STORE_NAME               1 (b)

assigns the rotated values to a and b respectively.

How Python's a, b = b, a works?

1 Answers1