How do I pass a variable by reference

Question

The Python documentation seems unclear about whether parameters are passed by reference or value  and the following code produces the unchanged value  Original   class PassByReference      def   init   self           self variable    Original          self change self variable          print self variable       def change self  var           var    Changed    Is there something I can do to pass the variable by actual reference

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You can merely use an empty class as an instance to store reference objects because internally object attributes are stored in an instance dictionary  See the example   class RefsObj object        A class which helps to create references to variables       pass         an example of usage def change ref var ref obj       ref obj val   24  ref obj   RefsObj   ref obj val   1 print ref obj val    or print ref obj val for python2 change ref var ref obj  print ref obj val

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Arguments are passed by assignment  The rationale behind this is twofold    the parameter passed in is actually a reference to an object  but the reference is passed by value  some data types are mutable  but others aren t   So    If you pass a mutable object into a method  the method gets a reference to that same object and you can mutate it to your heart s delight  but if you rebind the reference in the method  the outer scope will know nothing about it  and after you re done  the outer reference will still point at the original object   If you pass an immutable object to a method  you still can t rebind the outer reference  and you can t even mutate the object    To make it even more clear  let s have some examples    List - a mutable type  Let s try to modify the list that was passed to a method   def try to change list contents the list       print  got   the list      the list append  four       print  changed to   the list   outer list     one    two    three    print  before  outer list     outer list  try to change list contents outer list  print  after  outer list     outer list    Output   before  outer list     one    two    three   got   one    two    three   changed to   one    two    three    four   after  outer list     one    two    three    four     Since the parameter passed in is a reference to outer list  not a copy of it  we can use the mutating list methods to change it and have the changes reflected in the outer scope   Now let s see what happens when we try to change the reference that was passed in as a parameter   def try to change list reference the list       print  got   the list      the list     and    we    can    not    lie       print  set to   the list   outer list     we    like    proper    English    print  before  outer list     outer list  try to change list reference outer list  print  after  outer list     outer list    Output   before  outer list     we    like    proper    English   got   we    like    proper    English   set to   and    we    can    not    lie   after  outer list     we    like    proper    English     Since the the list parameter was passed by value  assigning a new list to it had no effect that the code outside the method could see  The the list was a copy of the outer list reference  and we had the list point to a new list  but there was no way to change where outer list pointed   String - an immutable type  It s immutable  so there s nothing we can do to change the contents of the string  Now  let s try to change the reference  def try to change string reference the string       print  got   the string      the string    In a kingdom by the sea      print  set to   the string   outer string    It was many and many a year ago   print  before  outer string     outer string  try to change string reference outer string  print  after  outer string     outer string    Output   before  outer string   It was many and many a year ago got It was many and many a year ago set to In a kingdom by the sea after  outer string   It was many and many a year ago   Again  since the the string parameter was passed by value  assigning a new string to it had no effect that the code outside the method could see  The the string was a copy of the outer string reference  and we had the string point to a new string  but there was no way to change where outer string pointed   I hope this clears things up a little   EDIT  It s been noted that this doesn t answer the question that  David originally asked   Is there something I can do to pass the variable by actual reference    Let s work on that   How do we get around this   As  Andrea s answer shows  you could return the new value  This doesn t change the way things are passed in  but does let you get the information you want back out   def return a whole new string the string       new string   something to do with the old string the string      return new string    then you could call it like my string   return a whole new string my string    If you really wanted to avoid using a return value  you could create a class to hold your value and pass it into the function or use an existing class  like a list   def use a wrapper to simulate pass by reference stuff to change       new string   something to do with the old string stuff to change 0       stuff to change 0    new string    then you could call it like wrapper    my string  use a wrapper to simulate pass by reference wrapper   do something with wrapper 0     Although this seems a little cumbersome

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While pass by reference is nothing that fits well into python and should be rarely used there are some workarounds that actually can work to get the object currently assigned to a local variable or even reassign a local variable from inside of a called function   The basic idea is to have a function that can do that access and can be passed as object into other functions or stored in a class   One way is to use global  for global variables  or nonlocal  for local variables in a function  in a wrapper function   def change wrapper       wrapper 7   x   5 def setter val       global x     x   val print x    The same idea works for reading and deleting a variable   For just reading there is even a shorter way of just using lambda  x which returns a callable that when called returns the current value of x  This is somewhat like  call by name  used in languages in the distant past   Passing 3 wrappers to access a variable is a bit unwieldy so those can be wrapped into a class that has a proxy attribute   class ByRef      def   init   self  r  w  d           self  read   r         self  write   w         self  delete   d     def set self  val           self  write val      def get self           return self  read       def remove self           self  delete       wrapped   property get  set  remove     left as an exercise for the reader  define set  get  remove as local functions using global   nonlocal r   ByRef get  set  remove  r wrapped   15   Pythons  reflection  support makes it possible to get a object that is capable of reassigning a name variable in a given scope without defining functions explicitly in that scope   class ByRef      def   init   self  locs  name           self  locs   locs         self  name   name     def set self  val           self  locs self  name    val     def get self           return self  locs self  name      def remove self           del self  locs self  name      wrapped   property get  set  remove   def change x       x wrapped   7  def test me        x   6     print x      change ByRef locals     x        print x    Here the ByRef class wraps a dictionary access  So attribute access to wrapped is translated to a item access in the passed dictionary  By passing the result of the builtin locals and the name of a local variable this ends up accessing a local variable  The python documentation as of 3 5 advises that changing the dictionary might not work but it seems to work for me

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A lot of insights in answers here  but i think an additional point is not clearly mentioned here explicitly    Quoting from python documentation https   docs python org 2 faq programming html what-are-the-rules-for-local-and-global-variables-in-python     In Python  variables that are only referenced inside a function are implicitly global  If a variable is assigned a new value anywhere within the function   s body  it   s assumed to be a local  If a variable is ever assigned a new value inside the function  the variable is implicitly local  and you need to explicitly declare it as    global     Though a bit surprising at first  a moment   s consideration explains this  On one hand  requiring global for assigned variables provides a bar against unintended side-effects  On the other hand  if global was required for all global references  you   d be using global all the time  You   d have to declare as global every reference to a built-in function or to a component of an imported module  This clutter would defeat the usefulness of the global declaration for identifying side-effects    Even when passing a mutable object to a function this still applies  And to me clearly explains the reason for the difference in behavior between assigning to the object and operating on the object in the function   def test l       print  Received   l   id l      l    0  0  0      print  Changed to   l  id l     New local object created  breaking link to global l  l   1 2 3  print  Original   l  id l  test l  print  After   l  id l    gives   Original  1  2  3  4454645632 Received  1  2  3  4454645632 Changed to  0  0  0  4474591928 After  1  2  3  4454645632   The assignment to an global variable that is not declared global therefore creates a new local object and breaks the link to the original object

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I am new to Python  started yesterday  though I have been programming for 45 years   I came here because I was writing a function where I wanted to have two so called out-parameters  If it would have been only one out-parameter  I wouldn t get hung up right now on checking how reference value works in Python  I would just have used the return value of the function instead  But since I needed two such out-parameters I felt I needed to sort it out  In this post I am going to show how I solved my situation  Perhaps others coming here can find it valuable  even though it is not exactly an answer to the topic question  Experienced Python programmers of course already know about the solution I used  but it was new to me  From the answers here I could quickly see that Python works a bit like Javascript in this regard  and that you need to use workarounds if you want the reference functionality  But then I found something neat in Python that I don t think I have seen in other languages before  namely that you can return more than one value from a function  in a simple comma separated way  like this  def somefunction p       a p 1     b p 2     c -p     return a  b  c  and that you can handle that on the calling side similarly  like this x  y  z   somefunction w   That was good enough for me and I was satisfied  No need to use some workaround  In other languages you can of course also return many values  but then usually in the from of an object  and you need to adjust the calling side accordingly  The Python way of doing it was nice and simple  If you want to mimic by reference even more  you could do as follows  def somefunction a  b  c       a   a   2     b   b   a     c   a   b   c     return a  b  c  x   3 y   5 z   10 print F quot Before    x    y    z  quot    x  y  z   somefunction x  y  z   print F quot After     x    y    z  quot    which gives this result  Before   3  5  10   After    6  11  660

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You got some really good answers here   x     2  4  4  5  5   print x    2  4  4  5  5  def go  li       li     5  6  7  8      re-assigning what li POINTS TO  does not     change the value of the ORIGINAL variable x  go  x    print x    2  4  4  5  5    STILL      raw input   press any key to continue

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The problem comes from a misunderstanding of what variables are in Python  If you re used to most traditional languages  you have a mental model of what happens in the following sequence   a   1 a   2   You believe that a is a memory location that stores the value 1  then is updated to store the value 2  That s not how things work in Python  Rather  a starts as a reference to an object with the value 1  then gets reassigned as a reference to an object with the value 2  Those two objects may continue to coexist even though a doesn t refer to the first one anymore  in fact they may be shared by any number of other references within the program   When you call a function with a parameter  a new reference is created that refers to the object passed in  This is separate from the reference that was used in the function call  so there s no way to update that reference and make it refer to a new object  In your example   def   init   self       self variable    Original      self Change self variable   def Change self  var       var    Changed    self variable is a reference to the string object  Original   When you call Change you create a second reference var to the object  Inside the function you reassign the reference var to a different string object  Changed   but the reference self variable is separate and does not change   The only way around this is to pass a mutable object  Because both references refer to the same object  any changes to the object are reflected in both places   def   init   self                self variable     Original       self Change self variable   def Change self  var       var 0     Changed

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A simple trick I normally use is to just wrap it in a list   def Change self  var       var 0     Changed   variable     Original   self Change variable        print variable 0     Yeah I know this can be inconvenient  but sometimes it is simple enough to do this

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Think of stuff being passed by assignment instead of by reference by value  That way  it is always clear  what is happening as long as you understand what happens during the normal assignment   So  when passing a list to a function method  the list is assigned to the parameter name  Appending to the list will result in the list being modified  Reassigning the list inside the function will not change the original list  since   a    1  2  3  b   a b append 4  b     a    b   print a  b        prints  1  2  3  4    a    b     Since immutable types cannot be modified  they seem like being passed by value - passing an int into a function means assigning the int to the function s parameter  You can only ever reassign that  but it won t change the original variables value

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Since it seems to be nowhere mentioned an approach to simulate references as known from e g  C   is to use an  update  function and pass that instead of the actual variable  or rather   name     def need to modify update       update 42    set new value 42       other code  def call it        value   21     def update value new value           nonlocal value         value   new value     need to modify update value      print value    prints 42   This is mostly useful for  out-only references  or in a situation with multiple threads   processes  by making the update function thread   multiprocessing safe    Obviously the above does not allow reading the value  only updating it

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I found the other answers rather long and complicated  so I created this simple diagram to explain the way Python treats variables and parameters

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There are no variables in Python  The key to understanding parameter passing is to stop thinking about  variables   There are names and objects in Python and together they appear like variables  but it is useful to always distinguish the three    Python has names and objects  Assignment binds a name to an object  Passing an argument into a function also binds a name  the parameter name of the function  to an object    That is all there is to it  Mutability is irrelevant to this question   Example   a   1   This binds the name a to an object of type integer that holds the value 1   b   x   This binds the name b to the same object that the name x is currently bound to  Afterward  the name b has nothing to do with the name x anymore   See sections 3 1 and 4 2 in the Python 3 language reference   How to read the example in the question  In the code shown in the question  the statement self Change self variable  binds the name var  in the scope of function Change  to the object that holds the value  Original  and the assignment var    Changed   in the body of function Change  assigns that same name again  to some other object  that happens to hold a string as well but could have been something else entirely    How to pass by reference  So if the thing you want to change is a mutable object  there is no problem  as everything is effectively passed by reference   If it is an immutable object  e g  a bool  number  string   the way to go is to wrap it in a mutable object  The quick-and-dirty solution for this is a one-element list  instead of self variable  pass  self variable  and in the function modify var 0    The more pythonic approach would be to introduce a trivial  one-attribute class  The function receives an instance of the class and manipulates the attribute

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As you can state you need to have a mutable object  but let me suggest you to check over the global variables as they can help you or even solve this kind of issue   http   docs python org 3 faq programming html what-are-the-rules-for-local-and-global-variables-in-python  example    gt  gt  gt  def x y           global z         z   y       gt  gt  gt  x  lt function x at 0x00000000020E1730 gt   gt  gt  gt  y Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  NameError  name  y  is not defined  gt  gt  gt  z Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  NameError  name  z  is not defined   gt  gt  gt  x 2   gt  gt  gt  x  lt function x at 0x00000000020E1730 gt   gt  gt  gt  y Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt  NameError  name  y  is not defined  gt  gt  gt  z 2

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edit - Blair has updated his enormously popular answer so that it is now accurate   I think it is important to note that the current post with the most votes  by Blair Conrad   while being correct with respect to its result  is misleading and is borderline incorrect based on its definitions   While there are many languages  like C  that allow the user to either pass by reference or pass by value  Python is not one of them   David Cournapeau s answer points to the real answer and explains why the behavior in Blair Conrad s post seems to be correct while the definitions are not   To the extent that Python is pass by value  all languages are pass by value since some piece of data  be it a  value  or a  reference   must be sent  However  that does not mean that Python is pass by value in the sense that a C programmer would think of it   If you want the behavior  Blair Conrad s answer is fine   But if you want to know the nuts and bolts of why Python is neither pass by value or pass by reference  read David Cournapeau s answer

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There is a little trick to pass an object by reference  even though the language doesn t make it possible  It works in Java too  it s the list with one item   -   class PassByReference      def   init   self  name           self name   name  def changeRef ref       ref 0    PassByReference  Michael    obj   PassByReference  Peter   print obj name  p    obj    A pointer to obj   -  changeRef p   print p 0  name   p- gt name   It s an ugly hack  but it works   -P

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Aside from all the great explanations on how this stuff works in Python  I don t see a simple suggestion for the problem  As you seem to do create objects and instances  the pythonic way of handling instance variables and changing them is the following   class PassByReference      def   init   self           self variable    Original          self Change           print self variable      def Change self           self variable    Changed    In instance methods  you normally refer to self to access instance attributes  It is normal to set instance attributes in   init   and read or change them in instance methods  That is also why you pass self als the first argument to def Change   Another solution would be to create a static method like this   class PassByReference      def   init   self           self variable    Original          self variable   PassByReference Change self variable          print self variable       staticmethod     def Change var           var    Changed          return var

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Technically  Python always uses pass by reference values  I am going to repeat my other answer to support my statement   Python always uses pass-by-reference values  There isn t any exception  Any variable assignment means copying the reference value  No exception  Any variable is the name bound to the reference value  Always   You can think about a reference value as the address of the target object  The address is automatically dereferenced when used  This way  working with the reference value  it seems you work directly with the target object  But there always is a reference in between  one step more to jump to the target   Here is the example that proves that Python uses passing by reference     If the argument was passed by value  the outer lst could not be modified  The green are the target objects  the black is the value stored inside  the red is the object type   the yellow is the memory with the reference value inside -- drawn as the arrow  The blue solid arrow is the reference value that was passed to the function  via the dashed blue arrow path   The ugly dark yellow is the internal dictionary   It actually could be drawn also as a green ellipse  The colour and the shape only says it is internal    You can use the id   built-in function to learn what the reference value is  that is  the address of the target object    In compiled languages  a variable is a memory space that is able to capture the value of the type  In Python  a variable is a name  captured internally as a string  bound to the reference variable that holds the reference value to the target object  The name of the variable is the key in the internal dictionary  the value part of that dictionary item stores the reference value to the target   Reference values are hidden in Python  There isn t any explicit user type for storing the reference value  However  you can use a list element  or element in any other suitable container type  as the reference variable  because all containers do store the elements also as references to the target objects  In other words  elements are actually not contained inside the container -- only the references to elements are

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Since dictionaries are passed by reference  you can use a dict variable to store any referenced values inside it     returns the result of adding numbers  a  and  b  def AddNumbers a  b  ref     using a dict for reference     result   a   b     ref  multi     a   b   reference the multi  ref  multi   is number     ref  msg      The result      str result      was nice     reference any string  errors  e t c   ref  msg   is string     return result   return the sum  number1   5 number2   10 ref        init a dict like that so it can save all the referenced values  this is because all dictionaries are passed by reference  while strings and numbers do not   sum   AddNumbers number1  number2  ref  print  sum     sum                the return value print  multi     ref  multi       a referenced value print  msg     ref  msg           a referenced value

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In this case the variable titled var in the method Change is assigned a reference to self variable  and you immediately assign a string to var  It s no longer pointing to self variable  The following code snippet shows what would happen if you modify the data structure pointed to by var and self variable  in this case a list    gt  gt  gt  class PassByReference          def   init   self               self variable     Original               self change self variable              print self variable                      def change self  var               var append  Changed         gt  gt  gt  q   PassByReference     Original    Changed    gt  gt  gt     I m sure someone else could clarify this further

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I used the following method to quickly convert a couple of Fortran codes to Python   True  it s not pass by reference as the original question was posed  but is a simple work around in some cases   a 0 b 0 c 0 def myfunc a b c       a 1     b 2     c 3     return a b c  a b c   myfunc a b c  print a b c

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Python   s pass-by-assignment scheme isn   t quite the same as C     s reference parameters option  but it turns out to be very similar to the argument-passing model of the C language  and others  in practice    Immutable arguments are effectively passed    by value     Objects such as integers and strings are passed by object reference instead of by copying  but because you can   t change immutable objects in place anyhow  the effect is much like making a copy  Mutable arguments are effectively passed    by pointer     Objects such as lists and dictionaries are also passed by object reference  which is similar to the way C passes arrays as pointers   mutable objects can be changed in place in the function  much like C arrays

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It is neither pass-by-value or pass-by-reference - it is call-by-object  See this  by Fredrik Lundh    http   effbot org zone call-by-object htm  Here is a significant quote          variables  names  are not objects  they cannot be denoted by other variables or referred to by objects     In your example  when the Change method is called--a namespace is created for it  and var becomes a name  within that namespace  for the string object  Original   That object then has a name in two namespaces  Next  var    Changed  binds var to a new string object  and thus the method s namespace forgets about  Original   Finally  that namespace is forgotten  and the string  Changed  along with it

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given the way python handles values and references to them  the only way you can reference an arbitrary instance attribute is by name   class PassByReferenceIsh      def   init   self           self variable    Original          self change  variable           print self variable      def change self  var           self   dict   var     Changed    in real code you would  of course  add error checking on the dict lookup

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Effbot  aka Fredrik Lundh  has described Python s variable passing style as call-by-object   http   effbot org zone call-by-object htm  Objects are allocated on the heap and pointers to them can be passed around anywhere      When you make an assignment such as x   1000  a dictionary entry is created that maps the string  x  in the current namespace to a pointer to the integer object containing one thousand     When you update  x  with x   2000  a new integer object is created and the dictionary is updated to point at the new object   The old one thousand object is unchanged  and may or may not be alive depending on whether anything else refers to the object   When you do a new assignment such as y   x  a new dictionary entry  y  is created that points to the same object as the entry for  x   Objects like strings and integers are immutable   This simply means that there are no methods that can change the object after it has been created   For example  once the integer object one-thousand is created  it will never change   Math is done by creating new integer objects  Objects like lists are mutable   This means that the contents of the object can be changed by anything pointing to the object   For example  x       y   x  x append 10   print y will print  10    The empty list was created   Both  x  and  y  point to the same list   The append method mutates  updates  the list object  like adding a record to a database  and the result is visible to both  x  and  y   just as a database update would be visible to every connection to that database     Hope that clarifies the issue for you

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Pass-By-Reference in Python is quite different from the concept of pass by reference in C   Java     Java amp C   primitive types include string pass by value copy   Reference type is passed by reference address copy  so all changes made in the parameter in the called function are visible to the caller  C    Both pass-by-reference or pass-by-value are allowed  If a parameter is passed by reference  you can either modify it or not depending upon whether the parameter was passed as const or not  However  const or not  the parameter maintains the reference to the object and reference cannot be assigned to point to a different object within the called function  Python   Python is    pass-by-object-reference     of which it is often said     Object references are passed by value     Read here 1  Both the caller and the function refer to the same object but the parameter in the function is a new variable which is just holding a copy of the object in the caller  Like C    a parameter can be either modified or not in function - This depends upon the type of object passed  eg  An immutable object type cannot be modified in the called function whereas a mutable object can be either updated or re-initialized  A crucial difference between updating or re-assigning re-initializing the mutable variable is that updated value gets reflected back in the called function whereas the reinitialized value does not  Scope of any assignment of new object to a mutable variable is local to the function in the python  Examples provided by  blair-conrad are great to understand this

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alternatively you could use ctypes witch would look something like this import ctypes  def f a       a value 2398    resign the value in a function  a   ctypes c int 0  print  quot pre f quot   a  f a  print  quot post f quot   a   as a is a c int and not a python integer and apperently passed by reference  however you have to be carefull as strange things could happen and is therefor not advised

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Here is the simple  I hope  explanation of the concept pass by object used in Python  Whenever you pass an object to the function  the object itself is passed  object in Python is actually what you d call a value in other programming languages  not the reference to this object  In other words  when you call   def change me list      list    1  2  3   my list    0  1  change me my list    The actual object -  0  1   which would be called a value in other programming languages  is being passed  So in fact the function change me will try to do something like    0  1     1  2  3    which obviously will not change the object passed to the function  If the function looked like this   def change me list      list append 2    Then the call would result in    0  1  append 2    which obviously will change the object  This answer explains it well

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Since your example happens to be object-oriented  you could make the following change to achieve a similar result   class PassByReference      def   init   self           self variable    Original          self change  variable           print self variable       def change self  var           setattr self  var   Changed      o variable will equal  Changed  o   PassByReference   assert o variable     Changed

[python] How do I pass a variable by reference?

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