Question

I’m trying to understand Python’s approach to variable scope. In this example, why is f() able to alter the value of x, as perceived within main(), but not the value of n?

def f(n, x):     n = 2     x.append(4)     print('In f():', n, x)  def main():     n = 1     x = [0,1,2,3]     print('Before:', n, x)     f(n, x)     print('After: ', n, x)  main() 

Output:

Before: 1 [0, 1, 2, 3] In f(): 2 [0, 1, 2, 3, 4] After:  1 [0, 1, 2, 3, 4] 

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See also:

• How do I pass a variable by reference?
• Are Python variables pointers? Or else, what are they?

Answer 1

Some answers contain the word "copy" in the context of a function call. I find it confusing.

Python doesn’t copy objects you pass during a function call ever.

Function parameters are names. When you call a function, Python binds these parameters to whatever objects you pass (via names in a caller scope).

Objects can be mutable (like lists) or immutable (like integers and strings in Python). A mutable object you can change. You can’t change a name, you just can bind it to another object.

Your example is not about scopes or namespaces, it is about naming and binding and mutability of an object in Python.

def f(n, x): # these `n`, `x` have nothing to do with `n` and `x` from main()     n = 2    # put `n` label on `2` balloon     x.append(4) # call `append` method of whatever object `x` is referring to.     print('In f():', n, x)     x = []   # put `x` label on `[]` ballon     # x = [] has no effect on the original list that is passed into the function 

Here are nice pictures on the difference between variables in other languages and names in Python.