lambda function don't closure the parameter in Python? [duplicate]

This question already has answers here: What do lambda function closures capture? (7 answers) 开发者_运维技巧 Closed 8 months ago.

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()

---

you need to do:

lambda i=i: pprint(i)

instead to capture the current value of i

---

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.

---

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.

---

The lambda functions create closures over the i variable. After the first for loop is finished, i has value 4.

Then the second for loop starts and all the lambda functions are executed. Each of them then prints the current value of i, which is 4.

---

The answer: Because the code inside the lambda is using your global variable i.

Your second variant will do the same as the first one with lambda if you remove the parameter i:

def closure():

Instead of

def closure(i):

I.e. the code inside the function will use the global variable i.