Consider the following code. ``python x = [0, 1, 2] x[0], x[2] = x[2], x[0] `` What does the second assignment do?

Question

x = [0, 1, 2]
x[0], x[2] = x[2], x[0]

Options

• AIt shortens the list.
• BIt reverses the list.
• CIt doesn't change the list.
• DIt extends the list.

• A
  2% (1)
• B
  81% (39)
• C
  10% (5)
• D
  6% (3)

Explanation

Option B is correct because x[0], x[2] = x[2], x[0] is a simultaneous swap: Python evaluates the right-hand side (x[2], x[0] → 2, 0) as a tuple before any assignment occurs, then assigns 2 to x[0] and 0 to x[2], transforming [0, 1, 2] into [2, 1, 0] - which is the list reversed. Option A (shortens) and D (extends) are wrong because tuple unpacking on the left side of = only reassigns existing indices, never adds or removes elements. Option C is wrong because the first and last elements genuinely change places - the list is mutated, not left intact.

Memory tip: Think of the right-hand side as a snapshot taken before anything moves - Python "photographs" both values simultaneously, so there's no accidental overwrite. Whenever you see a, b = b, a, that's always a clean swap.

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