[closures] Why doesn't the outer function argument need to be specified as nonlocal in the inner function?

Question

Python

Hey!
There is a closure:

def outer(x):
  count = 555
  num = 666
  def inner():
    nonlocal count, num
    return x
  return inner


muller = outer(777)
print(muller())
print(muller.__closure__[0].cell_contents, muller.__closure__[1].cell_contents, muller.__closure__[2].cell_contents)

As far as I understand, the variable x gets into __closure__ only when it is used in inner?
Why shouldn't it be specified as nonlocal along with count, num?

Answer 1

global and nonlocal must be specified if you are going to change the variable.
If you're just reading it, there's no need for it.