Why does Python sometimes not create new objects for function arguments?

Question

I assumed that whenever you passed an object into a function, it automatically created a separate copy for you that you could modify without changing the original. However, the following code terminates with this error every time.

>>> testvar = ['a' for a in range(100000)]
>>> def func(arg):
        while arg is testvar:
            arg.pop()
>>> func(testvar)
Traceback (most recent call last):
  File "<pyshell#47>", line 1, in <module>
    thing(m)
  File "<pyshell#43>", line 3, in thing
    obj.pop()
IndexError: pop from empty list

possible duplicate of In Python, why can a function modify some arguments as perceived by the caller, but not others? — TigerhawkT3
– TigerhawkT3, Commented Aug 7, 2015 at 19:26

chepner · Accepted Answer · 2015-08-07 20:50:30Z

2

"I assumed that whenever you passed an object into a function, it automatically created a separate copy for you that you could modify without changing the original."

Well, your assumption there is wrong. Variables are passed by reference, not value. This includes if you set a default value for an argument to an empty list, and change it in the function, the next time you try to use that default value, it won't be an empty list anymore.

def foo(bar=[]):
    bar.append("baz")

foo()
foo()
# bar is now ["baz", "baz"]

edited Aug 7, 2015 at 20:50

chepner

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answered Aug 7, 2015 at 19:32

fishsticks

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8 Comments

chepner Over a year ago

"Most" variables? I'd say all.

fishsticks Over a year ago

@chepner Not integers and floats, at least.

DSM Over a year ago

@fishsticks: integers behave exactly the same way as lists-- no copies are made. It's simply that integers aren't mutable. Try the_num = 1; print(id(the_num)); def add(num): print(id(num)); add(the_num).

chepner Over a year ago

num += 1 has the same effect as num = num + 1 because ints are immutable; the original object reference by num is still the same object that is passed as the argument.

thebjorn Over a year ago

@DSM in that particular case, it's because 1 is special (small integers are cached). You are correct in general though.

|

Alex Jadczak · Accepted Answer · 2015-08-07 19:44:58Z

2

from: https://docs.python.org/3.4/library/copy.html

Assignment statements in Python do not copy objects, they create bindings between a target and an object. For collections that are mutable or contain mutable items, a copy is sometimes needed so one can change one copy without changing the other. This module provides generic shallow and deep copy operations (explained below).

when you are calling your function, you are doing the following:

func(arg=testvar)

If you want to have an actual copy of your data you could modify your function to the following:

def func(arg):
    arg_copy = arg[:]
    # mutate arg_copy here without mutating arg

answered Aug 7, 2015 at 19:44

Alex Jadczak

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