Python functions call by reference

Question

In some languages you can pass a parameter by reference or value by using a special reserved word like ref or val  When you pass a parameter to a Python function it never alters the value of the parameter on leaving the function The only way to do this is by using the global reserved word  or as i understand it currently   Example 1  k   2  def foo  n        n   n   n      clarity regarding comment below      square   n      return square  j   foo k  print j print k  would show  gt  gt 4  gt  gt 2  showing k to be unchanged  In this example the variable n is never changed Example 2  n   0 def foo        global n     n   n   n     return n  In this example the variable n is changed  Is there any way in Python to call a function and tell Python that the parameter is either a value or reference parameter instead of using global  Secondly  in the A level Cambridge exams they now say a function returns a single value whereas a procedure returns more than one value  I was taught a function has a return statement and procedure does not  during the 80s  Why is this now incorrect

User · Answer

The answer given is  def set 4 x      y         for i in x         y append i     y 0    4    return y   and   l    0   def set 3 x        x 0    3  set 3 l  print l 0     which is the best answer so far as it does what it says in the question  However it does seem a very clumsy way compared to VB or Pascal Is it the best method we have   Not only is it clumsy  it involves mutating the original parameter in some way manually eg by changing the  original parameter to a list  or copying it to another list rather than just saying   use this parameter as a  value   or  use this  one as a  reference   Could the simple answer be there is no reserved word for this but these are great work arounds

User · Answer

Python is neither pass-by-value nor pass-by-reference   It s more of  object references are passed by value  as described here      Here s why it s not pass-by-value   Because  def append list       list append 1   list    0  reassign list  append list     returns  0 1  showing that some kind of reference was clearly passed as pass-by-value does not allow a function to alter the parent scope at all   Looks like pass-by-reference then  hu   Nope    Here s why it s not pass-by-reference  Because  def reassign list     list    0  1   list    0  reassign list  print list    returns  0  showing that the original reference was destroyed when list was reassigned   pass-by-reference would have returned  0 1    For more information look here    If you want your function to not manipulate outside scope  you need to make a copy of the input parameters that creates a new object   from copy import copy  def append list     list2   copy list    list2 append 1    print list2  list    0  append list  print list

User · Answer

class demoClass      x   4     y   3 foo1   demoClass   foo1 x   2 foo2   demoClass   foo2 y   5 def mySquare myObj       myObj x   myObj x  2     myObj y   myObj y  2 print  foo1 x     foo1 x  print  foo1 y     foo1 y  print  foo2 x     foo2 x  print  foo2 y     foo2 y  mySquare foo1  mySquare foo2  print  After square    print  foo1 x     foo1 x  print  foo1 y     foo1 y  print  foo2 x     foo2 x  print  foo2 y     foo2 y

User · Answer

Consider that the variable is a box and the value it points to is the  thing  inside the box   1  Pass by reference   function shares the same box and thereby the thing inside also   2  Pass by value   function creates a new box  a replica of the old one  including a copy of whatever thing is inside it  Eg  Java - functions create a copy of the box and the thing inside it which can be  a primitive   a reference to an object   note that the copied reference in the new box and the original both still point to the same object  here the reference IS the thing inside the box  not the object it is pointing to   3  Pass by object-reference  the function creates a box  but it encloses the same thing the initial box was enclosing  So in Python   a  if the thing inside said box is mutable  changes made will reflect back in the original box   eg  lists   b  if the thing is immutable  like python strings and numeric types   then the box inside the function will hold the same thing UNTIL you try to change its value  Once changed  the thing in the function s box is a totally new thing compared to the original one  Hence id   for that box will now give the identity of the new thing it encloses

User · Answer

Technically python do not pass arguments by value  all by reference  But     since python has two types of objects  immutable and mutable  here is what happens   Immutable arguments are effectively passed by value  string  integer  tuple are all immutable object types  While they are technically  quot passed by reference quot   like all parameters   since you can t change them in-place inside the function it looks behaves as if it is passed by value   Mutable arguments are effectively passed by reference  lists or dictionaries are passed by its pointers  Any in-place change inside the function like  append or del  will affect the original object    This is how Python is designed  no copies and all are passed by reference  You can explicitly pass a copy  def sort array         do sort     return array  data    1  2  3  sort data       here you passed a copy  Last point I would like to mention which is a function has its own scope  def do any stuff to these objects a  b        a   a   2      del b  last name    number   1   immutable hashmap     first name     john    last name    legend     mutable do any stuff to these objects number  hashmap   print number    1   oh  it should be 2   no a is changed inisde the function scope print hashmap      first name    john

User · Answer

Hope the following description sums it up well   There are two things to consider here - variables and objects     If you are passing a variable  then it s pass by value  which means the changes made to the variable within the function are local to that function and hence won t be reflected globally  This is more of a  C  like behavior    Example   def changeval  myvar       myvar   20      print  values inside the function     myvar    return  myvar   10  changeval  myvar    print  values outside the function     myvar   O P   values inside the function   20  values outside the function   10    If you are passing the variables packed inside a mutable object  like a list  then the changes made to the object are reflected globally as long as the object is not re-assigned    Example   def changelist  mylist       mylist2   a       mylist append mylist2      print  values inside the function     mylist    return  mylist    1 2 3   changelist  mylist    print  values outside the function     mylist   O P   values inside the function    1  2  3    a    values outside the function    1  2  3    a       Now consider the case where the object is re-assigned  In this case  the object refers to a new memory location which is local to the function in which this happens and hence not reflected globally     Example    def changelist  mylist       mylist   a       print  values inside the function     mylist    return  mylist    1 2 3   changelist  mylist    print  values outside the function     mylist   O P   values inside the function     a   values outside the function    1  2  3

User · Answer

Python already call by ref     let s take example     def foo var         print hex id var        x   1   any value   print hex id x      I think the id   give the ref       foo x    OutPut     0x50d43700  with you might give another hex number deppend on your memory    0x50d43700

User · Answer

OK  I ll take a stab at this  Python passes by object reference  which is different from what you d normally think of as  by reference  or  by value   Take this example   def foo x       print x  bar    some value  foo bar    So you re creating a string object with value  some value  and  binding  it to a variable named bar  In C  that would be similar to bar being a pointer to  some value    When you call foo bar   you re not passing in bar itself  You re passing in bar s value  a pointer to  some value   At that point  there are two  pointers  to the same string object   Now compare that to   def foo x       x    another value      print x  bar    some value  foo bar    Here s where the difference lies  In the line   x    another value    you re not actually altering the contents of x  In fact  that s not even possible  Instead  you re creating a new string object with value  another value   That assignment operator  It isn t saying  overwrite the thing x is pointing at with the new value   It s saying  update x to point at the new object instead   After that line  there are two string objects   some value   with bar pointing at it  and  another value   with x pointing at it    This isn t clumsy  When you understand how it works  it s a beautifully elegant  efficient system

User · Answer

So this is a little bit of a subtle point  because while Python only passes variables by value  every variable in Python is a reference  If you want to be able to change your values with a function call  what you need is a mutable object  For example   l    0   def set 3 x       x 0    3  set 3 l  print l 0     In the above code  the function modifies the contents of a List object  which is mutable   and so the output is 3 instead of 0   I write this answer only to illustrate what  by value  means in Python  The above code is bad style  and if you really want to mutate your values you should write a class and call methods within that class  as MPX suggests

User · Answer

In Python the passing by reference or by value has to do with what are the actual objects you are passing So if you are passing a list for example then you actually make this pass by reference since the list is a mutable object Thus you are passing a pointer to the function and you can modify the object  list   in the function body   When you are passing a string this passing is done by value so a new string object is being created and when the function terminates it is destroyed  So it all has to do with mutable and immutable objects

User · Answer

You can not change an immutable object  like str or tuple  inside a function in Python  but you can do things like   def foo y     y 0    y 0   2  x    5  foo x  print x 0     prints 25   That is a weird way to go about it  however  unless you need to always square certain elements in an array   Note that in Python  you can also return more than one value  making some of the use cases for pass by reference less important   def foo x  y      return x  2  y  2  a   2 b   3 a  b   foo a  b     a    4  b    9   When you return values like that  they are being returned as a Tuple which is in turn unpacked   edit  Another way to think about this is that  while you can t explicitly pass variables by reference in Python  you can modify the properties of objects that were passed in  In my example  and others  you can modify members of the list that was passed in  You would not  however  be able to reassign the passed in variable entirely  For instance  see the following two pieces of code look like they might do something similar  but end up with different results   def clear a x     x       def clear b x     while x  x pop    z    1 2 3  clear a z    z will not be changed clear b z    z will be emptied

User · Answer

There are essentially three kinds of  function calls    Pass by value Pass by reference Pass by object reference  Python is a PASS-BY-OBJECT-REFERENCE programming language  Firstly  it is important to understand that a variable  and the value of the variable  the object  are two seperate things  The variable  points to  the object  The variable is not the object  Again  THE VARIABLE IS NOT THE OBJECT Example  in the following line of code   gt  gt  gt  x          is the empty list  x is a variable that points to the empty list  but x itself is not the empty list  Consider the variable  x  in the above case  as a box  and  the value  of the variable      as the object inside the box  PASS BY OBJECT REFERENCE  Case in python   Here   quot Object references are passed by value  quot  def append one li       li append 1  x    0  append one x  print x  Here  the statement x    0  makes a variable x  box  that points towards the object  0   On the function being called  a new box li is created  The contents of li are the SAME as the contents of the box x  Both the boxes contain the same object  That is  both the variables point to the same object in memory  Hence  any change to the object pointed at by li will also be reflected by the object pointed at by x  In conclusion  the output of the above program will be   0  1   Note  If the variable li is reassigned in the function  then li will point to a separate object in memory  x however  will continue pointing to the same object in memory it was pointing to earlier  Example  def append one li       li    0  1  x    0  append one x  print x  The output of the program will be   0   PASS BY REFERENCE  The box from the calling function is passed on to the called function  Implicitly  the contents of the box  the value of the variable  is passed on to the called function  Hence  any change to the contents of the box in the called function will be reflected in the calling function  PASS BY VALUE  A new box is created in the called function  and copies of contents of the box from the calling function is stored into the new boxes  Hope this helps

[python] Python functions call by reference

Examples related to python