For loop evaluation order

I was looking into the all and any functions to see their guarantees about order.

For example, the documentation states that the any function is equivalent to:

def any(iterable):
    for element in iterable:
        if element:
            return True
    return False

However, when running the following code:

a = 1
any([a == 1, a[1] == 1])

I would expect the second expression to not be evaluated and, thus, would return True. Instead, the second line throws TypeError: 'int' object is not subscriptable. I initially believed that this was some inconsistency in the documentation but, indeed, when running with the for loop version, the result is the same.

My question is, how does it throw the exception, if it should not evaluate the second expression by returning True early? I thought that python only checked those errors when running the expressions.

Changing to

a = 1
any([a == 1, print('hello world')])

Also evaluates the print in both versions. Can anyone shine some light on this? Thanks in advance.

Note: using python 3.8.10 and interaction was performed in the python shell.

>Solution :

As @Barmar said, all the elements inside of the list must be evaluated when the list is created, to create something similar to that but lazy you could do:

def it():
    yield print(1)
    yield print(2)
    yield True
    yield print(3)
    
print(any(it()))

With output:

1
2
True