lambda function don't closure the parameter in Python?

Question

Code talks more:

from pprint import pprint

li = []

for i in range(5):
        li.append(lambda : pprint(i))

for k in li:
        k()

yield:

4
4
4
4
4

why not

0
1
2
3
4

??

Thanks.

P.S. If I write the complete decorator, it works as expected:

from pprint import pprint

li = []

#for i in range(5):
        #li.append(lambda : pprint(i))

def closure(i):
        def _func():
                pprint(i)
        return _func

for i in range(5):
        li.append(closure(i))

for k in li:
        k()

Answer 1

you need to do:

lambda i=i: pprint(i)

instead to capture the current value of i

Answer 2

It does properly reference i, only the thing is, that by the time your list gets populated, i would have the value assigned form the last item in the sequence when the iteration ended, so that's why your seeing 4 all over.

Answer 3

If you don't want to use a default argument -- which could introduce bugs if accidentally called with an argument -- you can use a nested lambda:

from pprint import pprint

li = []

for i in range(5):
    li.append((lambda x: lambda: pprint(x))(i))

for k in li:
    k()

This is an anonymous version of your closure function.