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How do I pass a variable by reference?

I wrote this class for testing:

class PassByReference:
 def __init__(self):
 self.variable = 'Original'
 self.change(self.variable)
 print(self.variable)
 def change(self, var):
 var = 'Changed'

When I tried creating an instance, the output was Original. So it seems like parameters in Python are passed by value. Is that correct? How can I modify the code to get the effect of pass-by-reference, so that the output is Changed?

_{Sometimes people are surprised that code like x = 1, where x is a parameter name, doesn't impact on the caller's argument, but code like x[0] = 1 does. This happens because item assignment and slice assignment are ways to mutate an existing object, rather than reassign a variable, despite the = syntax. See Why can a function modify some arguments as perceived by the caller, but not others? for details.}

_{See also What's the difference between passing by reference vs. passing by value? for important, language-agnostic terminology discussion.}

Answer*

Arguments are [passed by assignment][1]. 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

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](#986031)'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.


 [1]: http://docs.python.org/3/faq/programming.html#how-do-i-write-a-function-with-output-parameters-call-by-reference

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78

-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.

Ethan Furman
– Ethan Furman

2012年01月07日 06:41:59 +00:00
Commented Jan 7, 2012 at 6:41
5

"...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.

Honest Abe
– Honest Abe

2014年02月10日 23:47:43 +00:00
Commented Feb 10, 2014 at 23:47
3

@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".

Happy Ahmad
– Happy Ahmad

2018年07月09日 09:21:44 +00:00
Commented Jul 9, 2018 at 9:21
3

@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.

Blair Conrad
– Blair Conrad

2018年08月04日 09:57:37 +00:00
Commented Aug 4, 2018 at 9:57
3

@HappyAhmad it absolutely is different to call by reference. If Python supported call by reference, you could do something like def foo(&var): var = 2 then x = 0; y = 1; foo(x); foo(y) then print(x, y) would print 2 2

juanpa.arrivillaga
– juanpa.arrivillaga

2021年03月13日 19:58:51 +00:00
Commented Mar 13, 2021 at 19:58

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How do I pass a variable by reference?

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