/Python/How to catch stop itteration in generator as None

How to catch stop itteration in generator as None

# creating the list
L = [1,2,3,4,5,6]
# defining the generator
def generator(l):
    for i in l: yield i
    while True: yield  None
        

g = generator(L)

print(next(g)) # 1
print(next(g)) # 2
print(next(g)) # 3
print(next(g)) # 4
print(next(g)) # 5
print(next(g)) # 6
print(next(g)) # None
print(next(g)) # None
print(next(g)) # None

Similar codes

0 vote306 views1 answer
Unpacking using List in Python
Python
0 vote313 views1 answer
Api view wrapper django
Python Django
3 votes290 views1 answer
Draw a single pixel using pygame
Python
4 votes285 views1 answer
Urlpatterns = [ path('SignUp/', views.SignupPage, name='user_data')\
Python
0 vote322 views1 answer
Exit() function
Python
0 vote275 views1 answer
Python turtle draw 3 concentric circles
Python
30 votes311 views4 answers
How to convert data type of a column in pandas
Python
6 votes304 views2 answers
Markdown image embed url
Python
0 vote270 views2 answers
Get ContentType with django get_model
Python
1 vote287 views1 answer
Python inheritance remove an attribute
Python