What is super() when subclassing a list? How do I access the underlying list directly?-CodePudding

class Foo(list):
    def bar(self):
        return super().__getitem__(slice(None))
    
    def baz(self):
        return super()

a = Foo([0, 1, 2, 3])
print(a.bar()) # [0, 1, 2, 3]
print(a.baz()) # <super: <class 'Foo'>, <Foo object>>

# a.bar() provides a copy of the underlying list as can be seen from the fact that each result of a.bar() has a different id
print(id(a.bar())) # id1
print(id(a.bar())) # id2 != id1

I recently had a use case where I needed to subclass list and needed to access the underlying list from within the subclass (Foo). I thought super() would provide the underlying list, but it didn't. Instead I had to do super().__getitem__(slice(None)) which provides a copy of the underlying list. How do I access the underlying list directly? And what am I missing in my understanding of super()?

Many thanks!

CodePudding user response：

Just use your instance, as it's your list!

class Foo(list):
    pass


a = Foo([0, 1, 2, 3])
# __str__ method works fine
print(a)  # [0, 1, 2, 3]

for element in a:
    print(element)  # 0 1 2 3