Why can a function modify some arguments as perceived by the caller but not others

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

User · Accepted Answer

Some answers contain the word  copy  in a 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  strings in Python   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

User · Answer

Python is a pure pass-by-value language if you think about it the right way   A python variable stores the location of an object in memory   The Python variable does not store the object itself   When you pass a variable to a function  you are passing a copy of the address of the object being pointed to by the variable     Contrasst these two functions  def foo x       x 0    5  def goo x       x        Now  when you type into the shell   gt  gt  gt  cow    3 4 5   gt  gt  gt  foo cow   gt  gt  gt  cow  5 4 5    Compare this to goo    gt  gt  gt  cow    3 4 5   gt  gt  gt  goo cow   gt  gt  gt  goo  3 4 5    In the first case  we pass a copy the address of cow to foo and foo modified the state of the object residing there   The object gets modified   In the second case you pass a copy of the address of cow to goo  Then goo proceeds to change that copy  Effect  none   I call this the pink house principle   If you make a copy of your address and tell a  painter to paint the house at that address pink  you will wind up with a pink house  If you give the painter a copy of your address and tell him to change it to a new address  the address of your house does not change   The explanation eliminates a lot of confusion   Python passes the addresses variables store by value

User · Answer

Python is copy by value of reference   An object occupies a field in memory  and a reference is associated with that object  but itself occupies a field in memory  And name value is associated with a reference  In python function  it always copy the value of the reference  so in your code  n is copied to be a new name  when you assign that  it has a new space in caller stack  But for the list  the name also got copied  but it refer to the same memory since you never assign the list a new value   That is a magic in python

User · Answer

n is an int  immutable   and a copy is passed to the function  so in the function you are changing the copy   X is a list  mutable   and a copy of the pointer is passed o the function so x append 4  changes the contents of the list   However  you you said x    0 1 2 3 4  in your function  you would not change the contents of x in main

User · Answer

It  s because a list is a mutable object  You  re not setting x to the value of  0 1 2 3   you  re defining a label to the object  0 1 2 3    You should declare your function f   like this   def f n  x None       if x is None          x

User · Answer

My general understanding is that any object variable  such as a list or a dict  among others  can be modified through its functions  What I believe you are not able to do is reassign the parameter - i e   assign it by reference within a callable function   That is consistent with many other languages   Run the following short script to see how it works   def func1 x  l1       x   5     l1 append  nonsense    y   10 list1     meaning   func1 y  list1  print y  print list1

User · Answer

I will rename variables to reduce confusion   n -  nf or nmain   x -  xf or xmain   def f nf  xf       nf   2     xf append 4      print  In f      nf  xf  def main        nmain   1     xmain    0 1 2 3      print  Before    nmain  xmain     f nmain  xmain      print  After     nmain  xmain  main     When you call the function f  the Python runtime makes a copy of xmain and assigns it to xf  and similarly assigns a copy of  nmain to nf   In the case of n  the value that is copied is 1   In the case of x the value that is copied is not the literal list  0  1  2  3    It is a reference to that list   xf and xmain are pointing at the same list  so when you modify xf you are also modifying xmain   If  however  you were to write something like       xf     foo    bar       xf append 4    you would find that xmain has not changed   This is because  in the line xf     foo    bar   you have change xf to point to a new list   Any changes you make to this new list will have no effects on the list that xmain still points to   Hope that helps    -

User · Answer

f doesn t actually alter the value of x  which is always the same reference to an instance of a list   Rather  it alters the contents of this list   In both cases  a copy of a reference is passed to the function  Inside the function    n gets assigned a new value  Only the reference inside the function is modified  not the one outside it  x does not get assigned a new value  neither the reference inside nor outside the function are modified  Instead  x   s value is modified    Since both the x inside the function and outside it refer to the same value  both see the modification  By contrast  the n inside the function and outside it refer to different values after n was reassigned inside the function

User · Answer

I had modified my answer tons of times and realized i don t have to say anything  python had explained itself already   a    string  a replace  t        print a   gt  gt  gt   string   a   a replace  t        print a   gt  gt  gt   s ring   b   100 b   1 print b   gt  gt  gt  100  b   b   1 print b   gt  gt  gt  101    def test id arg       c   id arg      arg   123     d   id arg      return  a    test ids  b   id a  test id a  e   id a     b   c    e    d     this function do change original value del change like mutable arg       arg append 1      arg insert 0  9      arg remove 2      return  test 1    1  2  3  change like mutable test 1       this function doesn t  def wont change like str arg        arg    1  2  3       return   test 2    1  1  1  wont change like str test 2  print  Doesn t change like a imutable   test 2     This devil is not the reference   value   mutable or not   instance  name space or variable   list or str   IT IS THE SYNTAX  EQUAL SIGN

User · Answer

You ve got a number of answers already  and I broadly agree with J F  Sebastian  but you might find this useful as a shortcut   Any time you see varname    you re creating a new name binding within the function s scope   Whatever value varname was bound to before is lost within this scope   Any time you see varname foo   you re calling a method on varname   The method may alter varname  e g  list append    varname  or  rather  the object that varname names  may exist in more than one scope  and since it s the same object  any changes will be visible in all scopes    note that the global keyword creates an exception to the first case

User · Answer

If the functions are re-written with completely different variables and we call id on them  it then illustrates the point well  I didn t get this at first and read jfs  post with the great explanation  so I tried to understand convince myself   def f y  z       y   2     z append 4      print   In f                   id y   id z    def main        n   1     x    0 1 2 3      print   Before in main    n  x id n  id x       f n  x      print   After in main    n  x id n  id x    main   Before in main  1  0  1  2  3    94635800628352 139808499830024 In f                             94635800628384 139808499830024 After in main  1  0  1  2  3  4  94635800628352 139808499830024   z and x have the same id  Just  different tags for the same underlying structure as the article says

[python] Why can a function modify some arguments as perceived by the caller, but not others?

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