Why does `list.__iadd__` modify both original and assigned lists, while `list.__add__` only modifies the assigned list?

Behavior Discrepancy between list.__iadd__ and list.__add__

Consider the following snippet:

<code class="python">x = y = [1, 2, 3, 4]
x += [4]
print(x)  # Outputs: [1, 2, 3, 4, 4]
print(y)  # Outputs: [1, 2, 3, 4, 4]</code>

In contrast, observe this:

<code class="python">x = y = [1, 2, 3, 4]
x = x + [4]
print(x)  # Outputs: [1, 2, 3, 4, 4]
print(y)  # Outputs: [1, 2, 3, 4]</code>

Why do these two code snippets behave differently?

Explanation

The key difference lies in the use of the " " operator. In the first snippet:

• x = [4] uses the __iadd__ method of the list class, which:

  • Mutates the original list (x) in-place
  • Extends it with the elements from the second list ([4])

This results in both x and y being modified to include the value 4.

In the second snippet, however, x = x [4] uses the __add__ method:

• x [4] creates a new list by concatenating x with [4]
• The x variable is then reassigned to this new list
• y remains unchanged since it is a different object

Therefore, x has the value [1, 2, 3, 4, 4] while y still holds the original value [1, 2, 3, 4].

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