Invalid closure in Python's lambda function

Question

Consider this short code snippet:

class X:
    pass

xs = []
for s in ("one", "two", "three"):
    x = X()
    x.f = lambda: print(s)
    xs.append(x)

for x in xs:
    x.f()

It ouputs:

three
three
three

Is this really an expected behavior? Could you explain why this is not:

one
two
three

Possible duplicate of [Python lambda closure scoping](https://stackoverflow.com/questions/13355233/python-lambda-closure-scoping) — MisterMiyagi, Jun 28 '19 at 11:54

score 2 · Accepted Answer · answered Jun 28 '19 at 11:51

Your lambda function holds reference to s, hence the last assigned value to s is printed when called outside that for loop. Try the below code for your expected behaviour. Here a copy of that existing reference s is created in v as function argument and that value is printed inside the function f.

class X:
    pass

xs = []
for s in ("one", "two", "three"):
    x = X()
    def f(v=s): print(v)
    x.f = f
    xs.append(x)

for x in xs:
    x.f()

Output:

one
two
three

You can even exchange `def f(v=s): print(v)` with `x.f = lambda v=s: print(v)` — kalehmann, Jun 28 '19 at 11:56

score 2 · Answer 2 · answered Jun 28 '19 at 11:52

2

It happens because the s variable is a reference, not a value. And the value by reference will be resolved when it will be called, not created. To resolve the value in the creation time use the default argument.

lambda s=s: print(s)

answered Jun 28 '19 at 11:52

MartenCatcher

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Invalid closure in Python's lambda function

2 Answers2