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## parallel assigment in python
a, b = 2,3
print  a  #2
print b   #2

## This is because internally its a tuple unpacking

a = 1,2
print a    ## tuple packing

## Tuple unpacking
a,b = 3, 4
print a  #3
print b   #4 

## Python swapping
>>> a = 1
>>> b  = 2
>>> a,b  = b, a
>>> print a
2
>>> print b
1

"""

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Python separates the right-hand side expression from the left-hand side assignment. First the right-hand side is evaluated, and the result is stored on the stack, and then the left-hand side names are assigned using opcodes that take values from the stack again.

For tuple assignments with 2 or 3 items, Python just uses the stack directly:
"""

def swap():
  a, b = b, a
  
>>> dis.dis(swap)
  2           0 LOAD_FAST                0 (b)
              3 LOAD_FAST                1 (a)
              6 ROT_TWO
              7 STORE_FAST               1 (a)
             10 STORE_FAST               0 (b)
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE
"""

Python separates the right-hand side expression from the left-hand side assignment. First the right-hand side is evaluated, and the result is stored on the stack, and then the left-hand side names are assigned using opcodes that take values from the stack again.

For tuple assignments with 2 or 3 items, Python just uses the stack directly:

>>> import dis
>>> def foo(a, b):
...     a, b = b, a
... 
>>> dis.dis(foo)
  2           0 LOAD_FAST                1 (b)
              3 LOAD_FAST                0 (a)
              6 ROT_TWO             
              7 STORE_FAST               0 (a)
             10 STORE_FAST               1 (b)
             13 LOAD_CONST               0 (None)
             16 RETURN_VALUE      
""""             
After the two LOAD_FAST opcodes (which push a value from a variable onto the stack), the top of stack holds [a, b]. The ROT_TWO opcode swaps the top two positions on the stack so the stack now has [b, a] at the top. The two STORE_FAST opcodes then takes those two values and store them in the names on the left-hand side of the assignment. The first STORE_FAST pops a value of the top of the stack and puts it into a, the next pops again, storing the value in b. The rotation is needed because Python guarantees that assignments in a target list on the left-hand side are done from left to right.

For a 3-name assignment, ROT_THREE followed by ROT_TWO is executed to reverse the top three items on the stack.

For longer left-hand-side assignments, an explicit tuple is built:
"""
>>> def bar(a, b, c, d):
...     d, c, b, a = a, b, c, d
... 
>>> dis.dis(bar)
  2           0 LOAD_FAST                0 (a)
              3 LOAD_FAST                1 (b)
              6 LOAD_FAST                2 (c)
              9 LOAD_FAST                3 (d)
             12 BUILD_TUPLE              4
             15 UNPACK_SEQUENCE          4
             18 STORE_FAST               3 (d)
             21 STORE_FAST               2 (c)
             24 STORE_FAST               1 (b)
             27 STORE_FAST               0 (a)
             30 LOAD_CONST               0 (None)
             33 RETURN_VALUE