Does this way of swapping values in list use extra memory in Python?

Question

The usual way to swap values in a list is to use a temporary variable.

temp = l[i]
l[i] = l[j]
l[j] = temp

But in python you can do this:

l[i], l[j] = l[j], l[i]

How does this second method work? Is it the exact same process? Does it use less / more memory?

Answer 1

import dis


def swap_using_temp(a, b):
    temp = a
    a = b
    b = temp


def swap_using_unpacking(a, b):
    a, b = b, a

swap_using_unpacking does not require extra memory.

Explanation: If you disassemble the code using dis module of both the function described then you will see that in swap_using_unpacking there is a bytecode instruction ROT_TWO which swaps the 2 topmost elements of the stack(which don't require a third variable hence no extra memory is consumed).

---

dis.dis(swap_using_unpacking)

 11           0 LOAD_FAST                1 (b)
              2 LOAD_FAST                0 (a)
              4 ROT_TWO
              6 STORE_FAST               0 (a)
              8 STORE_FAST               1 (b)
             10 LOAD_CONST               0 (None)
             12 RETURN_VALUE

---

dis.dis(swap_using_temp)

  5           0 LOAD_FAST                0 (a)
              2 STORE_FAST               2 (temp)

  6           4 LOAD_FAST                1 (b)
              6 STORE_FAST               0 (a)

  7           8 LOAD_FAST                2 (temp)
             10 STORE_FAST               1 (b)
             12 LOAD_CONST               0 (None)
             14 RETURN_VALUE