By David Sykes

2009-06-12 10:23:51 8 Comments

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'

    def change(self, var):
        var = 'Changed'

Is there something I can do to pass the variable by actual reference?


@Alok Garg 2019-01-22 19:59:41

Pass-By-Reference in Python is quite different from the concept of pass by reference in C++/Java.

  • Java&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.

@Daniel Jour 2019-05-10 00:37:33

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
    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.

@Liakos 2019-05-05 14:14:56

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

@Lutz Prechelt 2014-02-11 11:29:07

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.

  1. Python has names and objects.
  2. Assignment binds a name to an object.
  3. 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 for this question.


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. Afterwards, the name b has nothing to do with the name x any more.

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

(edit 2019-04-28)

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.

@Ned Batchelder 2014-06-23 21:53:13

"Python has no variables" is a silly and confusing slogan, and I really wish people would stop saying it... :( The rest of this answer is good!

@Lutz Prechelt 2014-06-25 07:30:38

It may be shocking, but it is not silly. And I don't think it is confusing either: It hopefully opens up the recipient's mind for the explanation that is coming and puts her in a useful "I wonder what they have instead of variables" attitude. (Yes, your mileage may vary.)

@Ned Batchelder 2014-06-25 11:09:48

would you also say that Javascript has no variables? They work the same as Python's. Also, Java, Ruby, PHP, .... I think a better teaching technique is, "Python's variables work differently than C's."

@Ned Batchelder 2014-10-29 16:29:37

Yes, Java has variables. So does Python, and JavaScript, Ruby, PHP, etc. You wouldn't say in Java that int declares a variable, but Integer does not. They both declare variables. The Integer variable is an object, the int variable is a primitive. As an example, you demonstrated how your variables work by showing a = 1; b = a; a++ # doesn't modify b. That's exactly true in Python also (using += 1 since there is no ++ in Python)!

@Lutz Prechelt 2015-10-09 10:51:43

The concept of "variable" is complex and often vague: A variable is a container for a value, identified by a name. In Python, the values are objects, the containers are objects (see the problem?) and the names are actually separate things. I believe it is much tougher to get an accurate understanding of variables in this manner. The names-and-objects explanation appears more difficult, but is actually simpler.

@Sherlock70 2016-06-14 12:31:38

I really don't see how this answer could be helpful in light of the question. No solution is given, only the status quo of python is recited. How can you pass an argument, so that it can be changed? Most of the answers here don't give a solution. But this one, denies the existence of the obvious in the first lines, so it stands out of the pack. Sorry for the downvote, but it just made me a bit mad.

@Erich 2017-08-24 20:44:31

var·i·a·ble ˈverēəb(ə)l/Submit adjective 1. not consistent or having a fixed pattern; liable to change. I would argue that python does have "Variables"

@rkachach 2017-08-30 17:40:12

"That is all there is to it. Mutability is irrelevant for this question." yes it does. If the self.variable were mutable then it could be changed inside the function self.change(). Thus, passing an immutable type as parameter function will never affect the passed "variable".

@Christian Groleau 2017-11-22 21:43:37

I like this answer. To further the idea of "variables" vs "names" (emphasis on the quotes) may be found here: Code Like a Pythonista: Idiomatic Python

@sergzach 2018-05-03 14:14:11

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."


# 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

@Jesse Hogan 2017-09-10 02:19:53

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'

    def change(self, var):
        setattr(self, var, 'Changed')

# o.variable will equal 'Changed'
o = PassByReference()
assert o.variable == 'Changed'

@Bishwas Mishra 2018-03-14 11:48:30

Although this works. It is not pass by reference. It is 'pass by object reference'.

@Blair Conrad 2009-06-12 11:18:27

Arguments are passed by assignment. The rationale behind this is twofold:

  1. the parameter passed in is actually a reference to an object (but the reference is passed by value)
  2. some data types are mutable, but others aren't


  • 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)
    print('changed to', the_list)

outer_list = ['one', 'two', 'three']

print('before, outer_list =', outer_list)
print('after, outer_list =', outer_list)


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)
print('after, outer_list =', outer_list)


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)
print('after, outer_string =', outer_string)


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]


Although this seems a little cumbersome.

@Andrea Ambu 2009-06-12 11:52:33

Then the same is in C, when you pass "by reference" you're actually passing by value the reference... Define "by reference" :P

@Blair Conrad 2009-06-12 12:09:22

I'm not sure I understand your terms. I've been out of the C game for a while, but back when I was in it, there was no "pass by reference" - you could pass things, and it was always pass by value, so whatever was in the parameter list was copied. But sometimes the thing was a pointer, which one could follow to the piece of memory (primitive, array, struct, whatever), but you couldn't change the pointer that was copied from the outer scope - when you were done with the function, the original pointer still pointed to the same address. C++ introduced references, which behaved differently.

@David Cournapeau 2009-06-12 14:41:10

@andrea, not it is not like C at all. Conrad is correct, but the terms reference/values are confusing in python. That's why you should really use another term (see my link to effbot for a good explanation)

@Zac Bowling 2010-12-22 02:46:01

bring up mutability is just adding confusion. it actually doesn't have anything to with the issue. you state it correct at first in that you can change the reference for something not in your scope (like a pointer pointer).

@Cam Jackson 2011-09-06 02:30:01

@Zac Bowling Mutability is totally relevant. If strings were mutable, then the first string example would have a different output. It's important to know that setting the value of a passed-in string will create a new string object, not modify the passed one. Mutability is what prevents a string argument from behaving the same as say, an integer argument in this case.

@Zac Bowling 2011-09-08 17:54:35

@Cam Jackson actually it's not relevant. Passing by reference means giving the ability to change the pointer that the calling reference was using. Talking about changing the data the destination of that pointer adds confusion and just an implementation detail.

@Cam Jackson 2011-09-08 23:50:34

@Zac Bowling I don't really get how what you're saying is relevant, in a practical sense, to this answer. If a Python newcomer wanted to know about passing by ref/val, then the takeaway from this answer is: 1- You can use the reference that a function receives as its arguments, to modify the 'outside' value of a variable, as long as you don't reassign the parameter to refer to a new object. 2- Assigning to an immutable type will always create a new object, which breaks the reference that you had to the outside variable.

@Mark Ransom 2011-11-15 16:46:32

@CamJackson, you need a better example - numbers are also immutable objects in Python. Besides, wouldn't it be true to say that any assignment without subscripting on the left side of the equals will reassign the name to a new object whether it is immutable or not? def Foo(alist): alist = [1,2,3] will not modify the contents of the list from the callers perspective.

@Mark Ransom 2011-11-15 16:54:07

@BlairConrad, passing "by reference" in C is just a commonly recognized convention whereby you pass the pointer to something rather than the value of something. In that respect it behaves exactly like Python: you may update the value pointed to by the pointer but updating the pointer itself has no effect outside the function, since the pointer was passed by value.

@John Doe 2011-12-11 19:28:07

Cue the sys._getframe(1).f_locals['new_string'] = 'foo' hack. Or just using ctypes.. :)

@Ethan Furman 2012-01-07 06:41:59

-1. The code shown is good, the explanation as to how is completely wrong. See the answers by DavidCournapeau or DarenThomas for correct explanations as to why.

@Johan Lundberg 2012-01-30 20:52:54

"..., 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." This is wrong or at least not relevant - you can never rebind a reference.

@Lennart Regebro 2013-05-03 08:10:21

Oh, that's sad. "It's passed by value, but the value is a reference". Sigh. Well, everything is a value. When the value is a reference, that's pass by reference. The difference in Python is minimal as Pythons variables aren't like C's variables, but calling it pass by value is definitely incorrect.

@tom 2013-06-11 23:17:32

I down voted because the code(where you do assignment in function body) given for string and list is basically the same and has the same behavior. And list is mutable a string is not. Your answer did not cleared a thing for me. It just made me realize that I totally don't understand Python.

@Richard Fung 2013-11-17 18:31:47

@Lennart Regebro, just because the value is a reference to something doesn't make it pass by reference. In C++ you can actually pass by reference, and when you do, if you reassign the reference, the parameter you passed will actually be modified. In Python, if you try to reassign the parameter you will just be changing the mapping of the name in the namespace of your function, something which is fundamentally different. Still, the whole idea of pass by reference is so screwed up now because everyone is using it to describe a different thing.

@Lennart Regebro 2013-11-17 19:24:56

@Synderesis: "In C++"... - Well, yeah, but now we are talking about Python. Python is passing references. Since it is passing reference, that's reasonably "pass by reference". "In Python, if you try to reassign the parameter" - Yes, but that's a completely different issue. This is because Python doesn't have variables that point to a memory location like C/C++. It has objects, and variables are names for these objects.

@Lennart Regebro 2013-11-27 09:39:19

(Hence my upvote for David's post calling it "call-by-object".)

@Richard Fung 2013-11-29 22:26:30

@Lennart Regebro: I realize that, but in the end you could also consider Python to be pass by value with the values being references. Of course, that is why I added that last line anyway. Defining terms to mean different things for different languages is by all means useless, because the whole point of defining those terms is to make it clear what you are saying.

@Lennart Regebro 2013-11-30 08:55:51

This answer is hence factually incorrect, and have been upvotes as "thanks" by newbies who don't understand the internals of Python. Which is why it makes me sad.

@Richard Fung 2013-12-05 20:11:49

@Lennart Regebro "it is a reference passed as a value the type is a reference/pointer and you would have to dereference it to access it." I don't see how that is true. Java uses references but is pass by value, and behaves identically to Python in that it has objects and variables are names referencing these objects. The statement you made earlier "When the value is a reference, that's pass by reference." is incorrect, because if it's pass by value, you can reassign the reference. However, in pass by reference, if you try to reassign the variable you are actually modifying it.

@Lennart Regebro 2013-12-05 20:27:48

@Synderesis: Once again, then: See Davids answer. With your definition of pass-by-value, then EVERYTHING IS PASS BY VALUE. There exists no pass-by-reference with that wording of it, since pass-by-reference is when the value you pass is a reference, but you don't have to treat the value as a reference inside the function. What Java calls "pass-by-value, and the value is a reference" is exactly what pass-by-reference is. The whole point is that what Python and Java does does not exactly fit either concept as used in other languages.

@Richard Fung 2013-12-08 10:42:06

@Lennart Regebro "There exists no pass-by-reference with that wording of it, since pass-by-reference is when the value you pass is a reference" In C++ you can have true pass by reference. Java is pass by value:

@Lennart Regebro 2013-12-08 13:01:04

@Synderesis: I wish the "move this to chat" option was always available, coomments are really not the place for me to repeat the same thing over and over.

@Honest Abe 2014-02-10 23:47:43

"...parameter passed in..." is incorrect use of terminology. A parameter is a named entity in a function (or method) definition that specifies an argument (or in some cases, arguments) that the function can accept. An argument is a value passed to a function (or method) when calling the function.

@Elazar 2016-09-04 22:58:50

Unless you look at locals() etc., this is exactly like Java. In Java the address-space of references is an implementation detail, but in Python is a string in some scope (__dict__). In C it would be numbers in a global scope. Also, there is only one type of primitive in Python (reference to object) and several primitives in Java, one of which is reference to Object. Pass-by-reference in Python would mean "pass the scope and the name of the variable" but currently there is no such thing. it can be done explicitly - and it is, when emulating this mechanism using a class (which is a scope).

@Tommy 2016-09-07 18:52:28

I've most often seen the term "call-by-object-reference" not "by assignment"

@Bill 2016-12-31 22:47:38

As a means to pass a reference to a name, why not pass the name itself as a string? Then use eval(name) to get the current object that the name references.

@akki 2017-05-19 11:26:38

@rkachach 2017-08-30 17:31:16

"but if you rebind the reference in the method, the outer scope will know nothing about it", well the if the passed reference is mutable and you performed some operations (let say appended some values to the list) then these operations will be visible by the outer scope. The outer-scope will stop seeing changes only once the rebind is performed. All the previous operations do affect the original object.

@Ahmad 2018-07-09 09:21:44

@EthanFurman You are wrong. Even if you read the link in DavidCournapeau's answer that is Call By Object, you will see that all of the expressions “call by object”, “call by sharing“ or “call by object reference“, have the same meaning that is using an address to access the object. And for sure the address has to be determined from namespace. I can't accept that "Call by object reference" is different from "Call by reference".

@GeorgeOfTheRF 2018-08-03 08:44:50

@BlairConrad "The parameter passed in is actually a reference to an object (but the reference is passed by value)". This sounds similar to pass by reference. How is pass by reference and assignment different? Can you give an example to explain "pass by assignment"?

@Blair Conrad 2018-08-04 09:57:37

@ML_Pro, "pass by assignment" seems to be a term made up by the Python documenters to describe "pass by value". For the user, I see no functional difference between passing a value that is a reference (or handle) as happens in languages such as Java or C# and what Python does, and I'd never use the term "pass by assignment"; it was edited into the answer, I assume to align with the documentation.

@Mark Ransom 2011-11-15 17:45:28

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'

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']

def Change(self, var):
    var[0] = 'Changed'

@Cam Jackson 2011-11-16 00:03:53

Good succinct explanation. Your paragraph "When you call a function..." is one of the best explanations I've heard of the rather cryptic phrase that 'Python function parameters are references, passed by value.' I think if you understand that paragraph alone, everything else kind of just makes sense and flows as a logical conclusion from there. Then you just have to be aware of when you're creating a new object and when you're modifying an existing one.

@Glassjawed 2012-05-07 01:10:29

But how can you reassign the reference? I thought you can't change the address of 'var' but that your string "Changed" was now going to be stored in the 'var' memory address. Your description makes it seem like "Changed" and "Original" belong to different places in memory instead and you just switch 'var' to a different address. Is that correct?

@Mark Ransom 2012-05-07 01:46:48

@Glassjawed, I think you're getting it. "Changed" and "Original" are two different string objects at different memory addresses and 'var' changes from pointing to one to pointing to the other.

@Tim Richardson 2015-11-11 10:07:29

using the id() function helps clarify matters, because it makes it clear when Python creates a new object (so I think, anyway).

@Luke Davis 2017-01-10 02:37:27

Woah... so it makes sense that lists are mutable because one can imagine saying "ok python, make these boxes for me"... then the contents of the boxes can be modified. But things like strings and numbers can without complication point to a series of 1s and 0s in your memory, so it makes more sense to "tag" them with names.

@akki 2017-03-06 07:09:36

This answer and this blog post by effbot on python-objects together make things as clear as they can be.

@Tony Suffolk 66 2018-03-10 10:55:24

Function calls do not create new references - use the id function inside and outside of the function to confirm that. The difference is what happens to the object when you attempt to change it inside the function.

@Mark Ransom 2018-03-10 14:24:14

@TonySuffolk66 id gives the identity of the object referenced, not the reference itself.

@Tony Suffolk 66 2018-03-11 13:53:57

Mark - I took your use of reference to mean a object; when a function is called a new binding is created of an existing object to a new name (with a increase in the Ref count). Most documentation I have seen talks about this as a new binding, and not a new reference. Apologies for misunderstanding.

@Mark Ransom 2018-03-11 21:31:45

@TonySuffolk66 I come from a C++ background, so I might be a little loose on the terminology. A C++ reference is the closest thing to a Python variable.

@Tony Suffolk 66 2018-03-13 07:30:48

@MarkRansom - strictly speaking Python doesn't have 'variables' - well not in a C/C++ way of having a memory location which is known by that name. What Python has is one or more names which are bound to a reference to object - or sometime a name is bound to a container of object references. Oh the joys of terminology :-)

@Yasir Jan 2018-09-30 09:55:56

So, can we say that a mutable object is passed by reference and an immutable object is not ?

@Mark Ransom 2018-09-30 16:10:19

@YasirJan no, they're both passed by reference. It's just that for an immutable object the reference doesn't help you.

@Yasir Jan 2018-10-01 15:26:38

@MarkRansom Thanks. The 'call by assignment' way of describing it from python docs made more sense to me and I now I understand it.

@Minh Tran 2018-10-20 03:38:48

@MarkRansom What exactly is a reference? Your definition doesn't go beyond suggesting that it is "something" that points/refers to an object (which is a helpful definition to aid in describing what a reference does but not necessarily what it is).

@Mark Ransom 2018-10-20 04:21:58

@MinhTran in the simplest terms, a reference is something that "refers" to an object. The physical representation of that is most likely a pointer, but that's simply an implementation detail. It really is an abstract notion at heart.

@Charles Smith 2019-06-13 21:27:13

thank you for this Mark! Great, simple explanation

@mARK bLOORE 2016-10-31 15:33:23

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'
        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.

@textshell 2016-08-20 14:02:37

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):

x = 5
def setter(val):
    global x
    x = val

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):
    def get(self):
        return self._read()
    def remove(self):
    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
    change(ByRef(locals(), "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.

@Nuno Aniceto 2014-02-10 17:57:39

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!


>>> def x(y):
...     global z
...     z = y

>>> x
<function x at 0x00000000020E1730>
>>> y
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'y' is not defined
>>> z
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'z' is not defined

>>> x(2)
>>> x
<function x at 0x00000000020E1730>
>>> y
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'y' is not defined
>>> z

@Max P Magee 2014-06-30 18:39:20

I was tempted to post a similar response- the original questioner may not have known that what he wanted was in fact to use a global variable, shared among functions. Here's the link I would have shared:… In answer to @Tim, Stack Overflow is not only a question and answer site, it's a vast repository of reference knowledge that only gets stronger and more nuanced- much like an active wiki- with more input.

@Dolf Andringa 2016-02-07 23:18:36

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'
        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

    def Change(var):
        var = 'Changed'
        return var

@matino 2012-10-02 08:03:08

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]

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):

Then the call would result in:

[0, 1].append(2)

which obviously will change the object. This answer explains it well.

@pepr 2012-10-03 20:46:23

The problem is that the assignment does something else than you expect. The list = [1, 2, 3] causes reusing the list name for something else and forgeting the originally passed object. However, you can try list[:] = [1, 2, 3] (by the way list is wrong name for a variable. Thinking about [0, 1] = [1, 2, 3] is a complete nonsense. Anyway, what do you think means the object itself is passed? What is copied to the function in your opinion?

@Veky 2014-05-09 09:10:45

@pepr objects aren't literals. They are objects. The only way to talk about them is giving them some names. That's why it's so simple once you grasp it, but enormously complicated to explain. :-)

@pepr 2014-05-12 11:05:32

@Veky: I am aware of that. Anyway, the list literal is converted to the list object. Actually, any object in Python can exist without a name, and it can be used even when not given any name. And you can think about them as about anonymous objects. Think about objects being the elements of a lists. They need not a name. You can access them through indexing of or iterating through the list. Anyway, I insist on [0, 1] = [1, 2, 3] is simply a bad example. There is nothing like that in Python.

@Veky 2014-05-12 12:35:58

@pepr: I don't necessarily mean Python-definition names, just ordinary names. Of course alist[2] counts as a name of a third element of alist. But I think I misunderstood what your problem was. :-)

@Veky 2014-05-15 05:20:34

Argh. My English is obviously much worse than my Python. :-) I'll try just once more. I just said you have to give object some names just to talk about them. By that "names" I didn't mean "names as defined by Python". I know Python mechanisms, don't worry.

@Brad Porter 2016-08-08 16:46:05

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.

def myfunc(a,b,c):
    return a,b,c

a,b,c = myfunc(a,b,c)
print a,b,c

@Zenadix 2014-09-04 16:05:35

I found the other answers rather long and complicated, so I created this simple diagram to explain the way Python treats variables and parameters. enter image description here

@user22866 2016-01-03 15:02:44

lovely, makes it easy to spot the subtle diff that there is an intermediate assignment, not obvious to a casual onlooker. +1

@Martijn Pieters 2016-05-25 00:06:08

It doesn't matter if A is mutable or not. If you assign something different to B, A doesn't change. If an object is mutable, you can mutate it, sure. But that has nothing to do with assignment directly to a name..

@Zenadix 2016-05-25 16:19:27

@Martijn You're right. I removed the part of the answer that mentions mutability. I don't think it can get any simpler now.

@Martijn Pieters 2016-05-25 16:32:34

Thanks for the update, much better! What confuses most people is assignment to a subscription; e.g. B[0] = 2, vs. direct assignment, B = 2.

@Hatshepsut 2016-07-04 23:06:47

"A is assigned to B." Is that not ambiguous? I think in ordinary English that can mean either A=B or B=A.

@Madivad 2016-07-19 12:40:14

I do like the visual representation, but still misses the point of mutable vs immutable which makes the right leg moot since there will be no append available. (still got an upvote for the visual rep though) :)

@itmuckel 2016-04-21 16:47:42

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): = name

def changeRef(ref):
    ref[0] = PassByReference('Michael')

obj = PassByReference('Peter')

p = [obj] # A pointer to obj! ;-)

print p[0].name # p->name

It's an ugly hack, but it works. ;-P

@Minh Tran 2018-10-20 03:19:21

p is reference to a mutable list object which in turn stores the object obj. The reference 'p', gets passed into changeRef. Inside changeRef, a new reference is created (the new reference is called ref) that points to the same list object that p points to. But because lists are mutable, changes to the list are visible by both references. In this case, you used the ref reference to change the object at index 0 so that it subsequently stores the PassByReference('Michael') object. The change to the list object was done using ref but this change is visible to p.

@Minh Tran 2018-10-20 03:22:21

So now, the references p and ref point to a list object that stores the single object, PassByReference('Michael'). So it follows that p[0].name returns Michael. Of course, ref has now gone out of scope and may be garbage collected but all the same.

@Minh Tran 2018-10-20 03:25:29

You have not changed the private instance variable, name, of the original PassByReference object associated with the reference obj, though. In fact, will return Peter. The aforementioned comments assumes the definition Mark Ransom gave.

@Minh Tran 2018-10-20 03:32:11

Point being, I don't agree that it's a hack (which I take to mean to refer to something that works but for reasons unknown, untested, or unintended by the implementer). You simply replaced one PassByReference object with another PassByReference object in your list and referred to the latter of the two objects.

@KobeJohn 2010-06-27 12:33:45

(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.

@cayhorstmann 2012-12-20 03:49:01

It is simply not true that all languages are call by value. In C++ or Pascal (and surely many others that I don't know), you have call by reference. For example, in C++, void swap(int& x, int& y) { int temp = x; x = y; y = temp; } will swap the variables passed to it. In Pascal, you use var instead of &.

@KobeJohn 2015-10-09 15:03:12

I thought I had replied to this long ago but I don't see it. For completeness - cayhorstmann misunderstood my answer. I was not saying everything is call by value in the terms that most people first learn regarding C / C++. It was simply that some value is passed (value, name, pointer, etc.) and that the terms used in Blair's original answer were inaccurate.

@ajknzhol 2015-03-27 04:38:18

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.

@Joop 2014-09-12 14:40:18

A lot of insights in answers here, but i think an additional point is not clearly mentioned here explicitly. Quoting from python documentation

"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)
print "After", l, id(l)


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.

@pepr 2012-09-15 18:53:57

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:

Illustrated example of passing the argument

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.

@BugShotGG 2012-10-04 17:03:47

Actually this is confirmed its pass by reference value. +1 for this answer although the example wasnt good.

@cayhorstmann 2012-12-20 03:54:47

Inventing new terminology (such as "pass by reference value" or "call by object" is not helpful). "Call by (value|reference|name)" are standard terms. "reference" is a standard term. Passing references by value accurately describes the behavior of Python, Java, and a host of other languages, using standard terminology.

@pepr 2012-12-20 08:54:24

@cayhorstmann: The problem is that Python variable has not the same terminology meaning as in other languages. This way, call by reference does not fit well here. Also, how do you exactly define the term reference? Informally, the Python way could be easily described as passing the address of the object. But it does not fit with a potentially distributed implementation of Python.

@Honest Abe 2014-01-13 22:10:01

I like this answer, but you might consider if the example is really helping or hurting the flow. Also, if you replaced 'reference value' with 'object reference' you would be using terminology that we could consider 'official', as seen here: Defining Functions

@pepr 2014-01-14 23:31:13

Well, but the official says... "arguments are passed using call by value (where the value is always an object reference, not the value of the object)." This way, you may be tempted to substitute it textually as ... arguments are passed using call by reference, which is a bit confusing beacuse it is not true. The confusion is caused by a bit more complex situation where none of the classical terms fits perfectly. I did not find any simpler example that would illustrate the behaviour.

@Honest Abe 2014-01-16 06:28:53

There is a footnote indicated at the end of that quote, which reads: "Actually, call by object reference would be a better description, since if a mutable object is passed, the caller will see any changes the callee makes to it..." I agree with you that confusion is caused by trying to fit terminology established with other languages. Semantics aside, the things that need to be understood are: dictionaries / namespaces, name binding operations and the relationship of name→pointer→object (as you already know).

@Raymond Hettinger 2013-03-29 04:41:44

Effbot (aka Fredrik Lundh) has described Python's variable passing style as call-by-object:

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.

@Honest Abe 2014-01-13 22:21:16

I really appreciate learning about this from a developer. Is it true that the id() function returns the pointer's (object reference's) value, as pepr's answer suggests?

@Raymond Hettinger 2014-01-14 09:03:32

@HonestAbe Yes, in CPython the id() returns the address. But in other pythons such as PyPy and Jython, the id() is just a unique object identifier.

@David Cournapeau 2009-06-12 12:55:41

It is neither pass-by-value or pass-by-reference - it is call-by-object. See this, by Fredrik Lundh:

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.

@Luciano 2011-12-13 01:25:55

I find it hard to buy. To me is just as Java, the parameters are pointers to objects in memory, and those pointers are passed via the stack, or registers.

@David Cournapeau 2011-12-14 01:53:45

This is not like java. One of the case where it is not the same is immutable objects. Think about the trivial function lambda x: x. Apply this for x = [1, 2, 3] and x = (1, 2, 3). In the first case, the returned value will be a copy of the input, and identical in the second case.

@Mike Graham 2012-11-14 20:58:57

No, it's exactly like Java's semantics for objects. I'm not sure what you mean by "In the first case, the returned value will be a copy of the input, and identical in the second case." but that statement seems to be plainly incorrect.

@David Cournapeau 2012-11-16 06:34:14

My comment is indeed incorrect, but I stand on the fact that python semantics are different from java, as explained in the referred article. In python, whether changes within a function are visible outside it for its argument depends on the mutability, not on being a special type (say object vs primitive types in java), and the argument passing 'protocol' is the same for everything in python.

@cayhorstmann 2012-12-20 03:42:47

It is exactly the same as in Java. Object references are passed by value. Anyone who thinks differently should attach the Python code for a swap function that can swap two references, like this: a = [42] ; b = 'Hello'; swap(a, b) # Now a is 'Hello', b is [42]

@David Cournapeau 2012-12-21 10:00:41

Whether something is an object or not is irrelevant (especially since almost everything is an object in python, e.g. int, float, etc...). The mutability matters. You can write a swap function that would work for lists (mutable), but not for tuples or strings (immutable). See also this:, which clearly shows a different behavior than java in some cases.

@Claudiu 2013-07-17 18:59:25

It is exactly the same as Java when you pass objects in Java. However, Java also have primitives, which are passed by copying the value of the primitive. Thus they differ in that case.

@brain storm 2014-07-14 23:29:18

@Claudiu: Its not only in the case of Java primitives, any immutable object in Java will behave same as primitives such as String which is a Java object

@Elazar 2016-09-04 23:09:25

@Claudiu you are 100% right, except Java does not differ with the behavior of the additional primitives - it merely does have additional primitives, where Python don't. Other differences are visible when you look at locals() and some other meta-stuff - the scopes are implemented in a different way, and this implementation detail is visible in Python.

@bobobobo 2009-06-12 12:16:45

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' )

@laycat 2014-06-29 03:37:36

yea, however if you do x = [ 2, 4, 4, 5, 5], y = x, X[0] = 1 , print x # [1, 4 ,4, 5, 5] print y # [1, 4, 4, 5, 5]

@AmanicA 2011-08-05 22:52:00

A simple trick I normally use is to just wrap it in a list:

def Change(self, var):
    var[0] = 'Changed'

variable = ['Original']
print variable[0]

(Yeah I know this can be inconvenient, but sometimes it is simple enough to do this.)

@M Katz 2014-09-16 06:33:28

+1 for small amount of text giving the essential workaround to the problem of Python not having pass-by-reference. (As a follow-on comment/question that fits here as well as anywhere on this page: It's not clear to my why python can't provide a "ref" keyword like C# does, that simply wraps the caller's argument in a list like this, and treat references to the argument within the function as the 0th element of the list.)

@Justas 2015-02-01 05:49:17

Nice. To pass by ref, wrap in [ ]'s.

@Daren Thomas 2009-06-12 12:17:48

Think of stuff being passed by assignment instead of by reference/by value. That way, it is allways clear, what is happening as long as you understand what happens during 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 = ['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 functions parameter. You can only ever reassign that, but it won't change the originial variables value.

@Christian Groleau 2017-11-22 21:45:48

At first glance this answer seems to sidestep the original question. After a second read I've come to realize that this makes the matter quite clear. A good follow up to this "name assignment" concept may be found here: Code Like a Pythonista: Idiomatic Python

@Mike Mazur 2009-06-12 10:39:59

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:

>>> class PassByReference:
...     def __init__(self):
...         self.variable = ['Original']
...         self.change(self.variable)
...         print self.variable
...     def change(self, var):
...         var.append('Changed')
>>> q = PassByReference()
['Original', 'Changed']

I'm sure someone else could clarify this further.

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