How is `x = 42; x = lambda: x` parsed?

The variable x is created by the first assignment, and rebound with the second assignment.

Since the x in the lambda isn't evaluated until the lambda is called, calling it will evaluate to the most recently assigned value.

Note that this is not dynamic scoping - if it were dynamic, the following would print "99", but it prints "<function ...":

x = 42
x = lambda: x

def test(f):
  x = 99
  print(f())

test(x)

The first assignment is irrelevant; the x in the body of the lambda is bound late:

x = lambda: x # no need for a prior assignment
x = lambda: y # notice: no NameError occurs, *until it is called*

This is the same reason that creating lambdas in a loop is tricky, and is also used to make trees with the standard library defaultdict:

tree = lambda: defaultdict(tree)
t = tree()
t['foo']['bar']['baz'] = 'look ma, no intermediate steps'

Related