Hacker News Reader

Comment by theplague42 | Hacker News Reader

Assignment can be confusing already.

  >>> locals()['a'] = 1
  >>> a
  1

If anything, the walrus operator allows for tightly-scoped assignment, which is good in my opinion.

View on HN

You don't even need the locals() function to get into trouble:

    x = [1, 2, 3, 4]
    def foo():
        x[0] += 3  # Okay
    def bar():
        x += [3]   # UnboundLocalError
    def qux():
        x = [5, 6, 7, 8]  # Binds a new `x`.

    def bar():
        x += [3]   # UnboundLocalError

This is an especially funky one. x.extend([3]) would be allowed. Presumably x += [3] is not because it expands to x = x + [3]... However, the += operator on lists works the same as extend(), i.e. it changes the list in-place.

dis.dis(bar) shows:

              0 LOAD_FAST                0 (x)
              2 LOAD_CONST               1 (3)
              4 INPLACE_ADD
              6 STORE_FAST               0 (x)
              8 LOAD_CONST               0 (None)
             10 RETURN_VALUE

So INPLACE_ADD and STORE_FAST are essentially doing x = x.__iadd__([3])

Reply on news.ycombinator.com