Python class and global vs local variables

Question

I have a problem understanding what is happening with the outcome of the following pieces of code:

my_str = "outside func"
def func():
    my_str = "inside func"
    class C():
        print(my_str)
        print((lambda:my_str)())
        my_str = "inside C"
        print(my_str)

The output is:

outside func
inside func
inside C

Another piece of code is:

my_str = "not in class"
class C:
    my_str = "in the class"
    print([my_str for i in (1,2)])
    print(list(my_str for i in (1,2)))

The output is:

[‘in the class’, 'in the class’]
['not in class’, 'not in class’]

The question is:

1. What is happening here in each print() statement?
2. Can anyone explain why the print() statement get the string from the different namespaces?

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Edit 1:

I think this is different from this question because I humbly think the answers there do not explain this variation:

my_str = "outside func"
def func():
    my_str = "inside func"
    class C():
        print(my_str)
        print((lambda:my_str)())
       #my_str = "inside C"
        print(my_str)

The output is:

inside func
inside func
inside func

Edit 2:

Indeed, this is a duplicate from this question because as Martijn Pieters says:

The answer there states: If a name is assigned to within a class body, almost at the start. You assigned to my_str, making it the same case as y there. Commenting out that line means you are no longer assigning to my_str, making it the same case as x.

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Answer 1

There are four scopes here:

1. the global scope
2. the function scope
3. the class body
4. the lambda scope

When creating a class statement, the class body is executed as a function would and the local namespace of that 'function' is used as the class attributes.

However, a class body is not a scope, and as such behaves differently from functions. In a function body, binding defines scope; assign to a name in a function and it is marked as a local. In a class, assigning to a name makes it a global until you actually do the assignment.

So your first print() call finds my_str as a global, because later in the class body you bind to the name. This is a consequence of class bodies not taking part in scopes.

Next, you define a lambda and call it. my_str is never assigned to in that lambda so it has to be found in a parent scope. Here the class body is not a scope, and the next scope up is the function scope. This is why that line prints inside func.

Last, you assign to the local name my_str in the class body, and print that local.

The moment you remove the assignment, the names in the class are treated as non-local; the first and last print() statements are now equal, and the name is simply looked up according to the normal scoping rules.