Unpack python tuple with [ ]'s [duplicate]

Question

I know the canonical way to unpack a tuple is like this

a, b, c = (1, 2, 3)
# or
(a,b,c) = (1, 2, 3)

but noticed that you can unpack a tuple like this

[a, b, c] = (1, 2, 3)

Does the second method incur any extra cost due to some sort of cast or list construction? Is there a way to inspect how the python interpreter is dealing with this akin to looking at the assembly from a compiler?

Harborage

Answer 1

No, those are all exactly equivalent. One way to look at this empirically is to use the dis dissasembler:

>>> import dis
>>> dis.dis("a, b, c = (1, 2, 3)")
  1           0 LOAD_CONST               0 ((1, 2, 3))
              2 UNPACK_SEQUENCE          3
              4 STORE_NAME               0 (a)
              6 STORE_NAME               1 (b)
              8 STORE_NAME               2 (c)
             10 LOAD_CONST               1 (None)
             12 RETURN_VALUE
>>> dis.dis("(a, b, c) = (1, 2, 3)")
  1           0 LOAD_CONST               0 ((1, 2, 3))
              2 UNPACK_SEQUENCE          3
              4 STORE_NAME               0 (a)
              6 STORE_NAME               1 (b)
              8 STORE_NAME               2 (c)
             10 LOAD_CONST               1 (None)
             12 RETURN_VALUE
>>> dis.dis("[a, b, c] = (1, 2, 3)")
  1           0 LOAD_CONST               0 ((1, 2, 3))
              2 UNPACK_SEQUENCE          3
              4 STORE_NAME               0 (a)
              6 STORE_NAME               1 (b)
              8 STORE_NAME               2 (c)
             10 LOAD_CONST               1 (None)
             12 RETURN_VALUE
>>>

From the formal language specification, this is detailed here. This is part of the "target list", A relevant quote:

Assignment of an object to a target list, optionally enclosed in parentheses or square brackets, is recursively defined as follows....

Bonaventura

Nice explanations and dis analysis. – Pazia

Answer 2

Using godbolt and selecting Python as the language then entering the three lines of code, you can see they all have the same bytecode:

  1           0 LOAD_CONST               5 ((1, 2, 3))
              2 UNPACK_SEQUENCE          3
              4 STORE_NAME               0 (a)
              6 STORE_NAME               1 (b)
              8 STORE_NAME               2 (c)

  2          10 LOAD_CONST               6 ((1, 2, 3))
             12 UNPACK_SEQUENCE          3
             14 STORE_NAME               0 (a)
             16 STORE_NAME               1 (b)
             18 STORE_NAME               2 (c)

  3          20 LOAD_CONST               7 ((1, 2, 3))
             22 UNPACK_SEQUENCE          3
             24 STORE_NAME               0 (a)
             26 STORE_NAME               1 (b)
             28 STORE_NAME               2 (c)

So, they are the same, just different syntaxes.

Lawn

Answer 3

I have checked the execution time and number of cycles using Linux perf tools. All 3 programs are showing the same execution time, so I think there are not any extra costs.

Liegnitz