[fny]

`f` behaves just like closure. It can even be assigned to a variable.

    def f(x): x = 2 # <- this `x` is the same `x` as in the argument, it can be access via locals() internally

You can even assign `f`. For example, `function = f`.

Python is call by reference. Change my mind. `def f(x): x[0] = 1` will manipulate whatever object you pass to it.

[Jtsummers]

If Python were fully call by reference then:

  def g(y):
    y = 3
  x = 10
  g(x)
  x # => ??

If it's 3 then it's pass by reference here, if it's 10 it's pass by reference. Which is it?

Additionally, for something to be a closure it has to close over something (an environment). What does `f` or `g` close over? Note that they aren't changing any environment, they are "merely" functions, not closures. Python does have closures, but those aren't examples of them.

And being able to assign a function to a variable does not make a closure, or do you think that C has closures because it has function pointers?

[Jtsummers]

Oops:

If it's 3 then it's pass by reference here, if it's 10 it's pass by value. Which is it?

(Don't hit reply when you're rushing out the door.)

[kazinator]

> Python is call by reference. Change my mind.

No need to change your entire mind; just an incorrect definition of call-by-reference.

Passing a value which has reference semantics isn't call-by-reference.

Your f receives x by value. That value is a reference into a boxed object. The x[0] = 1 does not change x; it changes the boxed object.

  >>> def f(x):
  ...    x[0] = 1
  ... 
  >>> a = [ 0, 2, 3 ]
  >>> b = a
  >>> f(a)
  >>> a is b
  True

The f function can do nothing to make "a is b" false.

Under pass-by-reference, assigning to x would do that.

Under pass-by-reference, we can pass a reference to a, such that a can be replaced, without affecting b.

"pass-by {whatever}" refers to the semantics of the parameter and what it receives from the argument expression, and how, not the semantics of the argument object/value.