The mechanism behind multiple assignment in Python [duplicate]

Question

We all know multiple assignment can assign multiple variables at one time, and it is useful in swap. It works well in this situation:

nums = [2, 0, 1]
nums[0], nums[2] = nums[2], nums[0]
# nums=[1, 0, 2] directly, correct

, but it failed in more complex situation, such as:

nums = [2, 0, 1]
nums[0], nums[nums[0]] = nums[nums[0]], nums[0]
# nums=[1, 2, 1] directly, incorrect

nums = [2, 0, 1]
tmp = nums[0]
nums[0], nums[tmp] = nums[tmp], nums[0]
# nums=[1, 0, 2] with temporary variable, correct

It seems in nums[nums[0]], nums[0] will be assigned before, not at one time. it also failed in complex linklist node swap, such as:

cur.next, cur.next.next.next, cur.next.next = cur.next.next, cur.next, cur.next.next.next
# directly, incorrect

pre = cur.next
post = cur.next.next
cur.next, post.next, pre.next = post, pre, post.next
# with temporary variable, correct

So I want to know the mechanism behind multiple assignment in Python, and what is the Best Practices for this, temporary variable is the only way?

Answer 1

a, b = c, d

is equivalent to

temp = (c, d)
a = temp[0]  # Expression a is evaluated here, not earlier
b = temp[1]  # Expression b is evaluated here, not earlier

Personally I would recommend to write complex assignments explicitly with temporary variables as you showed it.

Another way is to carefully choose the order of the elements in the assignment:

nums[nums[0]], nums[0] = nums[0], nums[nums[0]]

Changes nums as you expect.