3438

Anyone tinkering with Python long enough has been bitten (or torn to pieces) by the following issue:

def foo(a=[]):
    a.append(5)
    return a

Python novices would expect this function called with no parameter to always return a list with only one element: [5]. The result is instead very different, and very astonishing (for a novice):

>>> foo()
[5]
>>> foo()
[5, 5]
>>> foo()
[5, 5, 5]
>>> foo()
[5, 5, 5, 5]
>>> foo()

A manager of mine once had his first encounter with this feature, and called it "a dramatic design flaw" of the language. I replied that the behavior had an underlying explanation, and it is indeed very puzzling and unexpected if you don't understand the internals. However, I was not able to answer (to myself) the following question: what is the reason for binding the default argument at function definition, and not at function execution? I doubt the experienced behavior has a practical use (who really used static variables in C, without breeding bugs?)

Edit:

Baczek made an interesting example. Together with most of your comments and Utaal's in particular, I elaborated further:

def a():
    print("a executed")
    return []

           
def b(x=a()):
    x.append(5)
    print(x)

a executed
>>> b()
[5]
>>> b()
[5, 5]

To me, it seems that the design decision was relative to where to put the scope of parameters: inside the function, or "together" with it?

Doing the binding inside the function would mean that x is effectively bound to the specified default when the function is called, not defined, something that would present a deep flaw: the def line would be "hybrid" in the sense that part of the binding (of the function object) would happen at definition, and part (assignment of default parameters) at function invocation time.

The actual behavior is more consistent: everything of that line gets evaluated when that line is executed, meaning at function definition.

25
  • 81
    Complementary question - Good uses for mutable default arguments Commented Feb 6, 2012 at 20:54
  • 11
    I have not doubt mutable arguments violate least astonishment principle for an average person, and I have seen beginners stepping there, then heroically replacing mailing lists with mailing tuples. Nevertheless mutable arguments are still in line with Python Zen (Pep 20) and falls into "obvious for Dutch" (understood/exploited by hard core python programmers) clause. The recommended workaround with doc string is the best, yet resistance to doc strings and any (written) docs is not so uncommon nowadays. Personally, I would prefer a decorator (say @fixed_defaults).
    – Serge
    Commented Apr 6, 2017 at 16:04
  • 6
    My argument when I come across this is: "Why do you need to create a function that returns a mutable that could optionally be a mutable you would pass to the function? Either it alters a mutable or creates a new one. Why do you need to do both with one function? And why should the interpreter be rewritten to allow you to do that without adding three lines to your code?" Because we are talking about rewriting the way the interpreter handles function definitions and evocations here. That's a lot to do for a barely necessary use case. Commented Jun 1, 2017 at 21:22
  • 32
    "Python novices would expect this function to always return a list with only one element: [5]." I'm a Python novice, and I wouldn't expect this, because obviously foo([1]) will return [1, 5], not [5]. What you meant to say is that a novice would expect the function called with no parameter will always return [5]. Commented Jul 6, 2017 at 16:08
  • 7
    This question asks "Why did this [the wrong way] get implemented so?" It doesn't ask "What's the right way?", which is covered by [Why does using arg=None fix Python's mutable default argument issue?]*(stackoverflow.com/questions/10676729/…). New users are almost always less interested in the former and much more in the latter, so that's sometimes a very useful link/dupe to cite.
    – smci
    Commented Apr 21, 2019 at 8:48

34 Answers 34

1934

Actually, this is not a design flaw, and it is not because of internals or performance. It comes simply from the fact that functions in Python are first-class objects, and not only a piece of code.

As soon as you think of it this way, then it completely makes sense: a function is an object being evaluated on its definition; default parameters are kind of "member data" and therefore their state may change from one call to the other - exactly as in any other object.

In any case, the effbot (Fredrik Lundh) has a very nice explanation of the reasons for this behavior in Default Parameter Values in Python. I found it very clear, and I really suggest reading it for a better knowledge of how function objects work.

23
  • 107
    To anyone reading the above answer, I strongly recommend you take the time to read through the linked Effbot article. As well as all the other useful info, the part on how this language feature can be used for result caching/memoisation is very handy to know! Commented Oct 14, 2011 at 0:05
  • 163
    Even if it's a first-class object, one might still envision a design where the code for each default value is stored along with the object and re-evaluated each time the function is called. I'm not saying that would be better, just that functions being first-class objects does not fully preclude it.
    – gerrit
    Commented Jan 11, 2013 at 10:55
  • 563
    Sorry, but anything considered "The biggest WTF in Python" is most definitely a design flaw. This is a source of bugs for everyone at some point, because no one expects that behavior at first - which means it should not have been designed that way to begin with. I don't care what hoops they had to jump through, they should have designed Python so that default arguments are non-static. Commented Jun 7, 2013 at 21:28
  • 323
    Whether or not it's a design flaw, your answer seems to imply that this behaviour is somehow necessary, natural and obvious given that functions are first-class objects, and that simply isn't the case. Python has closures. If you replace the default argument with an assignment on the first line of the function, it evaluates the expression each call (potentially using names declared in an enclosing scope). There is no reason at all that it wouldn't be possible or reasonable to have default arguments evaluated each time the function is called in exactly the same way.
    – Mark Amery
    Commented Jan 8, 2014 at 22:16
  • 64
    The design doesn't directly follow from functions are objects. In your paradigm, the proposal would be to implement functions' default values as properties rather than attributes.
    – bukzor
    Commented May 3, 2014 at 20:46
338

Suppose you have the following code

fruits = ("apples", "bananas", "loganberries")

def eat(food=fruits):
    ...

When I see the declaration of eat, the least astonishing thing is to think that if the first parameter is not given, that it will be equal to the tuple ("apples", "bananas", "loganberries")

However, suppose later on in the code, I do something like

def some_random_function():
    global fruits
    fruits = ("blueberries", "mangos")

then if default parameters were bound at function execution rather than function declaration, I would be astonished (in a very bad way) to discover that fruits had been changed. This would be more astonishing IMO than discovering that your foo function above was mutating the list.

The real problem lies with mutable variables, and all languages have this problem to some extent. Here's a question: suppose in Java I have the following code:

StringBuffer s = new StringBuffer("Hello World!");
Map<StringBuffer,Integer> counts = new HashMap<StringBuffer,Integer>();
counts.put(s, 5);
s.append("!!!!");
System.out.println( counts.get(s) );  // does this work?

Now, does my map use the value of the StringBuffer key when it was placed into the map, or does it store the key by reference? Either way, someone is astonished; either the person who tried to get the object out of the Map using a value identical to the one they put it in with, or the person who can't seem to retrieve their object even though the key they're using is literally the same object that was used to put it into the map (this is actually why Python doesn't allow its mutable built-in data types to be used as dictionary keys).

Your example is a good one of a case where Python newcomers will be surprised and bitten. But I'd argue that if we "fixed" this, then that would only create a different situation where they'd be bitten instead, and that one would be even less intuitive. Moreover, this is always the case when dealing with mutable variables; you always run into cases where someone could intuitively expect one or the opposite behavior depending on what code they're writing.

I personally like Python's current approach: default function arguments are evaluated when the function is defined and that object is always the default. I suppose they could special-case using an empty list, but that kind of special casing would cause even more astonishment, not to mention be backwards incompatible.

16
  • 55
    I think it's a matter of debate. You are acting on a global variable. Any evaluation performed anywhere in your code involving your global variable will now (correctly) refer to ("blueberries", "mangos"). the default parameter could just be like any other case. Commented Jul 15, 2009 at 18:16
  • 76
    Actually, I don't think I agree with your first example. I'm not sure I like the idea of modifying an initializer like that in the first place, but if I did, I'd expect it to behave exactly as you describe — changing the default value to ("blueberries", "mangos").
    – Ben Blank
    Commented Jul 15, 2009 at 18:26
  • 15
    The default parameter is like any other case. What is unexpected is that the parameter is a global variable, and not a local one. Which in turn is because the code is executed at function definition, not call. Once you get that, and that the same goes for classes, it's perfectly clear. Commented Jul 15, 2009 at 18:59
  • 29
    I find the example misleading rather than brilliant. If some_random_function() appends to fruits instead of assigning to it, the behaviour of eat() will change. So much for the current wonderful design. If you use a default argument that's referenced elsewhere and then modify the reference from outside the function, you are asking for trouble. The real WTF is when people define a fresh default argument (a list literal or a call to a constructor), and still get bit.
    – alexis
    Commented Oct 9, 2014 at 15:37
  • 31
    You just explicitly declared global and reassigned the tuple - there is absolutely nothing surprising if eat works differently after that. Commented Jan 26, 2015 at 16:07
317

The relevant part of the documentation:

Default parameter values are evaluated from left to right when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that the same “pre-computed” value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified. This is generally not what was intended. A way around this is to use None as the default, and explicitly test for it in the body of the function, e.g.:

def whats_on_the_telly(penguin=None):
    if penguin is None:
        penguin = []
    penguin.append("property of the zoo")
    return penguin
10
  • 300
    The phrases "this is not generally what was intended" and "a way around this is" smell like they're documenting a design flaw.
    – bukzor
    Commented May 3, 2014 at 20:53
  • 17
    @bukzor: Pitfalls need to be noted and documented, which is why this question is good and has received so many upvotes. At the same time, pitfalls don't necessarily need to be removed. How many Python beginners have passed a list to a function that modified it, and were shocked to see the changes show up in the original variable? Yet mutable object types are wonderful, when you understand how to use them. I guess it just boils down to opinion on this particular pitfall.
    – Matthew
    Commented Jun 19, 2014 at 17:54
  • 50
    The phrase "this is not generally what was intended" means "not what the programmer actually wanted to happen," not "not what Python is supposed to do."
    – holdenweb
    Commented Dec 19, 2014 at 11:48
  • 21
    @holdenweb Wow, I'm mega-late to the party. Given the context, bukzor is completely right: they're documenting behavior/consequence that was not "intended" when they they decided the language should exec the function's definition. Since it's an unintended consequence of their design choice, it's a design flaw. If it were not a design flaw, there'd be no need to even offer "a way around this".
    – code_dredd
    Commented Oct 3, 2017 at 7:35
  • 11
    We could take it to chat and discuss how else it could be, but the semantics have been thoroughly debated and nobody could come up with a sensible mechanism for create-default-value-on-call. One serious issue is that the scope on call is often entirely different from that on definition, making name resolution uncertain if defaults were evaluated at call time. A "way around" means "you can achieve your desired end in the following way," not "this is a mistake in Python's design."
    – holdenweb
    Commented Oct 3, 2017 at 16:03
149

I know nothing about the Python interpreter inner workings (and I'm not an expert in compilers and interpreters either) so don't blame me if I propose anything unsensible or impossible.

Provided that python objects are mutable I think that this should be taken into account when designing the default arguments stuff. When you instantiate a list:

a = []

you expect to get a new list referenced by a.

Why should the a=[] in

def x(a=[]):

instantiate a new list on function definition and not on invocation? It's just like you're asking "if the user doesn't provide the argument then instantiate a new list and use it as if it was produced by the caller". I think this is ambiguous instead:

def x(a=datetime.datetime.now()):

user, do you want a to default to the datetime corresponding to when you're defining or executing x? In this case, as in the previous one, I'll keep the same behaviour as if the default argument "assignment" was the first instruction of the function (datetime.now() called on function invocation). On the other hand, if the user wanted the definition-time mapping he could write:

b = datetime.datetime.now()
def x(a=b):

I know, I know: that's a closure. Alternatively Python might provide a keyword to force definition-time binding:

def x(static a=b):
7
  • 15
    You could do: def x(a=None): And then, if a is None, set a=datetime.datetime.now()
    – Anon
    Commented Jul 16, 2009 at 0:18
  • 36
    Thank you for this. I couldn't really put my finger on why this irks me to no end. You have done it beautifully with a minimum of fuzz and confusion. As someone comming from systems programming in C++ and sometimes naively "translating" language features, this false friend kicked me in the in the soft of the head big time, just like class attributes. I understand why things are this way, but I cannot help but dislike it, no matter what positive might come of it. At least it is so contrary to my experience, that I'll probably (hopefully) never forget it...
    – AndreasT
    Commented Apr 22, 2011 at 9:33
  • 7
    @Andreas once you use Python for long enough, you begin to see how logical it is for Python to interpret things as class attributes the way it does - it is only because of the particular quirks and limitations of languages like C++ (and Java, and C#...) that it makes any sense for contents of the class {} block to be interpreted as belonging to the instances :) But when classes are first-class objects, obviously the natural thing is for their contents (in memory) to reflect their contents (in code). Commented Jul 22, 2011 at 19:55
  • 9
    Normative structure is no quirk or limitation in my book. I know it can be clumsy and ugly, but you can call it a "definition" of something. The dynamic languages seem a bit like anarchists to me: Sure everybody is free, but you need structure to get someone to empty the trash and pave the road. Guess I'm old... :)
    – AndreasT
    Commented Jul 26, 2011 at 8:54
  • 10
    The function definition is executed at module load time. The function body is executed at function call time. The default argument is part of the function definition, not of the function body. (It gets more complicated for nested functions.) Commented Mar 30, 2015 at 11:28
100

Well, the reason is quite simply that bindings are done when code is executed, and the function definition is executed, well... when the functions is defined.

Compare this:

class BananaBunch:
    bananas = []

    def addBanana(self, banana):
        self.bananas.append(banana)

This code suffers from the exact same unexpected happenstance. bananas is a class attribute, and hence, when you add things to it, it's added to all instances of that class. The reason is exactly the same.

It's just "How It Works", and making it work differently in the function case would probably be complicated, and in the class case likely impossible, or at least slow down object instantiation a lot, as you would have to keep the class code around and execute it when objects are created.

Yes, it is unexpected. But once the penny drops, it fits in perfectly with how Python works in general. In fact, it's a good teaching aid, and once you understand why this happens, you'll grok python much better.

That said it should feature prominently in any good Python tutorial. Because as you mention, everyone runs into this problem sooner or later.

5
  • How do you define a class attribute that is different for each instance of a class?
    – Kieveli
    Commented Jul 15, 2009 at 19:04
  • 23
    If it's different for each instance it's not a class attribute. Class attributes are attributes on the CLASS. Hence the name. Hence they are the same for all instances. Commented Jul 15, 2009 at 19:17
  • 3
    How do you define an attribute in a class that is different for each instance of a class? (Re-defined for those who could not determine that a person not familiar with Python's naming convenctions might be asking about normal member variables of a class).
    – Kieveli
    Commented Jul 16, 2009 at 14:03
  • 6
    @Kievieli: You ARE talking about normal member variables of a class. :-) You define instance attributes by saying self.attribute = value in any method. For example __init__(). Commented Jul 16, 2009 at 14:14
  • @Kieveli: Two answers: you can't, because any thing you define at a class level will be a class attribute, and any instance that accesses that attribute will access the same class attribute; you can, /sort of/, by using propertys -- which are actually class level functions that act like normal attributes but save the attribute in the instance instead of the class (by using self.attribute = value as Lennart said). Commented Jan 7, 2012 at 4:45
92

Why don't you introspect?

I'm really surprised no one has performed the insightful introspection offered by Python (2 and 3 apply) on callables.

Given a simple little function func defined as:

>>> def func(a = []):
...    a.append(5)

When Python encounters it, the first thing it will do is compile it in order to create a code object for this function. While this compilation step is done, Python evaluates* and then stores the default arguments (an empty list [] here) in the function object itself. As the top answer mentioned: the list a can now be considered a member of the function func.

So, let's do some introspection, a before and after to examine how the list gets expanded inside the function object. I'm using Python 3.x for this, for Python 2 the same applies (use __defaults__ or func_defaults in Python 2; yes, two names for the same thing).

Function Before Execution:

>>> def func(a = []):
...     a.append(5)
...     

After Python executes this definition it will take any default parameters specified (a = [] here) and cram them in the __defaults__ attribute for the function object (relevant section: Callables):

>>> func.__defaults__
([],)

O.k, so an empty list as the single entry in __defaults__, just as expected.

Function After Execution:

Let's now execute this function:

>>> func()

Now, let's see those __defaults__ again:

>>> func.__defaults__
([5],)

Astonished? The value inside the object changes! Consecutive calls to the function will now simply append to that embedded list object:

>>> func(); func(); func()
>>> func.__defaults__
([5, 5, 5, 5],)

So, there you have it, the reason why this 'flaw' happens, is because default arguments are part of the function object. There's nothing weird going on here, it's all just a bit surprising.

The common solution to combat this is to use None as the default and then initialize in the function body:

def func(a = None):
    # or: a = [] if a is None else a
    if a is None:
        a = []

Since the function body is executed anew each time, you always get a fresh new empty list if no argument was passed for a.


To further verify that the list in __defaults__ is the same as that used in the function func you can just change your function to return the id of the list a used inside the function body. Then, compare it to the list in __defaults__ (position [0] in __defaults__) and you'll see how these are indeed refering to the same list instance:

>>> def func(a = []): 
...     a.append(5)
...     return id(a)
>>>
>>> id(func.__defaults__[0]) == func()
True

All with the power of introspection!


* To verify that Python evaluates the default arguments during compilation of the function, try executing the following:

def bar(a=input('Did you just see me without calling the function?')): 
    pass  # use raw_input in Py2

as you'll notice, input() is called before the process of building the function and binding it to the name bar is made.

3
  • 2
    Is id(...) needed for that last verification, or would the is operator answer the same question?
    – das-g
    Commented Mar 9, 2016 at 8:09
  • 3
    @das-g is would do just fine, I just used id(val) because I think it might be more intuitive. Commented Mar 9, 2016 at 8:20
  • Using None as the default severely limits the usefulness of the __defaults__ introspection, so I don't think that works well as a defense of having __defaults__ work the way it does. Lazy-evaluation would do more to keep function defaults useful from both sides.
    – Brilliand
    Commented Oct 18, 2019 at 5:32
72

I used to think that creating the objects at runtime would be the better approach. I'm less certain now, since you do lose some useful features, though it may be worth it regardless simply to prevent newbie confusion. The disadvantages of doing so are:

1. Performance

def foo(arg=something_expensive_to_compute())):
    ...

If call-time evaluation is used, then the expensive function is called every time your function is used without an argument. You'd either pay an expensive price on each call, or need to manually cache the value externally, polluting your namespace and adding verbosity.

2. Forcing bound parameters

A useful trick is to bind parameters of a lambda to the current binding of a variable when the lambda is created. For example:

funcs = [ lambda i=i: i for i in range(10)]

This returns a list of functions that return 0,1,2,3... respectively. If the behaviour is changed, they will instead bind i to the call-time value of i, so you would get a list of functions that all returned 9.

The only way to implement this otherwise would be to create a further closure with the i bound, ie:

def make_func(i): return lambda: i
funcs = [make_func(i) for i in range(10)]

3. Introspection

Consider the code:

def foo(a='test', b=100, c=[]):
   print a,b,c

We can get information about the arguments and defaults using the inspect module, which

>>> inspect.getargspec(foo)
(['a', 'b', 'c'], None, None, ('test', 100, []))

This information is very useful for things like document generation, metaprogramming, decorators etc.

Now, suppose the behaviour of defaults could be changed so that this is the equivalent of:

_undefined = object()  # sentinel value

def foo(a=_undefined, b=_undefined, c=_undefined)
    if a is _undefined: a='test'
    if b is _undefined: b=100
    if c is _undefined: c=[]

However, we've lost the ability to introspect, and see what the default arguments are. Because the objects haven't been constructed, we can't ever get hold of them without actually calling the function. The best we could do is to store off the source code and return that as a string.

8
  • 3
    you could achieve introspection also if for each there was a function to create the default argument instead of a value. the inspect module will just call that function.
    – yairchu
    Commented Jul 16, 2009 at 10:24
  • @SilentGhost: I'm talking about if the behaviour was changed to recreate it - creating it once is the current behaviour, and why the mutable default problem exists.
    – Brian
    Commented Jul 16, 2009 at 10:59
  • 4
    @yairchu: That assumes the construction is safe to so (ie has no side effects). Introspecting the args shouldn't do anything, but evaluating arbitrary code could well end up having an effect.
    – Brian
    Commented Jul 16, 2009 at 11:02
  • 5
    A different language design often just means writing things differently. Your first example could easily be written as: _expensive = expensive(); def foo(arg=_expensive), if you specifically don't want it reevaluated. Commented Jul 16, 2009 at 18:23
  • @Glenn - that's what I was referring to with "cache the variable externally" - it is a bit more verbose, and you end up with extra variables in your namespace though.
    – Brian
    Commented Jul 16, 2009 at 19:04
67

5 points in defense of Python

  1. Simplicity: The behavior is simple in the following sense: Most people fall into this trap only once, not several times.

  2. Consistency: Python always passes objects, not names. The default parameter is, obviously, part of the function heading (not the function body). It therefore ought to be evaluated at module load time (and only at module load time, unless nested), not at function call time.

  3. Usefulness: As Frederik Lundh points out in his explanation of "Default Parameter Values in Python", the current behavior can be quite useful for advanced programming. (Use sparingly.)

  4. Sufficient documentation: In the most basic Python documentation, the tutorial, the issue is loudly announced as an "Important warning" in the first subsection of Section "More on Defining Functions". The warning even uses boldface, which is rarely applied outside of headings. RTFM: Read the fine manual.

  5. Meta-learning: Falling into the trap is actually a very helpful moment (at least if you are a reflective learner), because you will subsequently better understand the point "Consistency" above and that will teach you a great deal about Python.

5
  • 24
    It took me a year to find this behavior is messing up my code on production, ended up removing a complete feature until I bumped into this design flaw by chance. I'm using Django. Since the staging environment did not have many requests, this bug never had any impact on QA. When we went live and received many simultaneous requests - some utility functions started overwriting each other's parameters! Making security holes, bugs and what not.
    – oriadam
    Commented Sep 5, 2015 at 13:09
  • 22
    @oriadam, no offense, but I wonder how you learned Python without running into this before. I am just learning Python now and this possible pitfall is mentioned in the official Python tutorial right alongside the first mention of default arguments. (As mentioned in point 4 of this answer.) I suppose the moral is—rather unsympathetically—to read the official docs of the language you use to create production software.
    – Wildcard
    Commented Aug 30, 2016 at 2:26
  • 2
    Also, it would be surprising (to me) if a function of unknown complexity was called in addition to the function call I am making.
    – Vatine
    Commented Sep 2, 2016 at 13:26
  • 8
    @oriadam, your company needs code review and actual expert coders in the language they write in by the time they have development, staging and production environments. Newbie bugs and bad code habits should not make it to production code Commented Jan 4, 2021 at 22:03
  • 2
    Senior coders cannot magically catch all bugs. Code review is excellent and necessary, but don't pretend they can prevent "newbie bugs" from making it to production. Even experts introduce such bugs themselves once in a while. Commented Dec 19, 2022 at 4:48
61

This behavior is easy explained by:

  1. function (class etc.) declaration is executed only once, creating all default value objects
  2. everything is passed by reference

So:

def x(a=0, b=[], c=[], d=0):
    a = a + 1
    b = b + [1]
    c.append(1)
    print a, b, c
  1. a doesn't change - every assignment call creates new int object - new object is printed
  2. b doesn't change - new array is build from default value and printed
  3. c changes - operation is performed on same object - and it is printed
3
  • (Actually, add is a bad example, but integers being immutable still is my main point.)
    – Anon
    Commented Jul 15, 2009 at 23:54
  • Realized it to my chagrin after checking to see that, with b set to [], b.__add__([1]) returns [1] but also leaves b still [] even though lists are mutable. My bad.
    – Anon
    Commented Jul 16, 2009 at 0:03
  • @ANon: there is __iadd__, but it doesn't work with int. Of course. :-)
    – Veky
    Commented May 8, 2014 at 13:16
44

1) The so-called problem of "Mutable Default Argument" is in general a special example demonstrating that:
"All functions with this problem suffer also from similar side effect problem on the actual parameter,"
That is against the rules of functional programming, usually undesiderable and should be fixed both together.

Example:

def foo(a=[]):                 # the same problematic function
    a.append(5)
    return a

>>> somevar = [1, 2]           # an example without a default parameter
>>> foo(somevar)
[1, 2, 5]
>>> somevar
[1, 2, 5]                      # usually expected [1, 2]

Solution: a copy
An absolutely safe solution is to copy or deepcopy the input object first and then to do whatever with the copy.

def foo(a=[]):
    a = a[:]     # a copy
    a.append(5)
    return a     # or everything safe by one line: "return a + [5]"

Many builtin mutable types have a copy method like some_dict.copy() or some_set.copy() or can be copied easy like somelist[:] or list(some_list). Every object can be also copied by copy.copy(any_object) or more thorough by copy.deepcopy() (the latter useful if the mutable object is composed from mutable objects). Some objects are fundamentally based on side effects like "file" object and can not be meaningfully reproduced by copy. copying

Example problem for a similar SO question

class Test(object):            # the original problematic class
  def __init__(self, var1=[]):
    self._var1 = var1

somevar = [1, 2]               # an example without a default parameter
t1 = Test(somevar)
t2 = Test(somevar)
t1._var1.append([1])
print somevar                  # [1, 2, [1]] but usually expected [1, 2]
print t2._var1                 # [1, 2, [1]] but usually expected [1, 2]

It shouldn't be neither saved in any public attribute of an instance returned by this function. (Assuming that private attributes of instance should not be modified from outside of this class or subclasses by convention. i.e. _var1 is a private attribute )

Conclusion:
Input parameters objects shouldn't be modified in place (mutated) nor they should not be binded into an object returned by the function. (If we prefere programming without side effects which is strongly recommended. see Wiki about "side effect" (The first two paragraphs are relevent in this context.) .)

2)
Only if the side effect on the actual parameter is required but unwanted on the default parameter then the useful solution is def ...(var1=None): if var1 is None: var1 = [] More..

3) In some cases is the mutable behavior of default parameters useful.

14
  • 7
    I hope you're aware that Python is not a functional programming language.
    – Veky
    Commented May 8, 2014 at 13:18
  • 11
    Yes, Python is a multi-paragigm language with some functional features. ("Don't make every problem look like a nail just because you have a hammer.") Many of them are in Python best practicies. Python has an interesting HOWTO Functional Programming Other features are closures and currying, not mentioned here.
    – hynekcer
    Commented May 8, 2014 at 15:54
  • 4
    I'd also add, at this late stage, that Python's assignment semantics have been designed explicitly to avoid data copying where necessary, so the creation of copies (and especially of deep copies) will affect both run-time and memory usage adversely. They should therefore be used only when necessary, but newcomers often have difficulty understanding when that is.
    – holdenweb
    Commented Jan 16, 2018 at 14:27
  • 1
    @holdenweb I agree. A temporary copy is the most usual way and sometimes the only possible way how to protect the original mutable data from an extraneous function that modifies them potentially. Fortunately a function that unreasonably modifies data is considered a bug and therefore uncommon.
    – hynekcer
    Commented Jan 17, 2018 at 21:41
  • 2
    @holdenweb My answer was about what interesting was missing in other answers. I hope that everybody who reads it, says: "Wow, a solution exists even for forgetful obsessed programmers even in more complicated cases, but I prefer ... I'm making a resolution to not modify parameters unexpectedly," Also you wrote in another comment "rebinding that name guarantees it can never be modified." (you meant rebinding also individual items) Diligence or copying is an acceptable price for freedom in Python and I like it.
    – hynekcer
    Commented Feb 19, 2018 at 10:12
43

What you're asking is why this:

def func(a=[], b = 2):
    pass

isn't internally equivalent to this:

def func(a=None, b = None):
    a_default = lambda: []
    b_default = lambda: 2
    def actual_func(a=None, b=None):
        if a is None: a = a_default()
        if b is None: b = b_default()
    return actual_func
func = func()

except for the case of explicitly calling func(None, None), which we'll ignore.

In other words, instead of evaluating default parameters, why not store each of them, and evaluate them when the function is called?

One answer is probably right there--it would effectively turn every function with default parameters into a closure. Even if it's all hidden away in the interpreter and not a full-blown closure, the data's got to be stored somewhere. It'd be slower and use more memory.

4
  • 10
    It wouldn't need to be a closure - a better way to think of it would simply to make the bytecode creating defaults the first line of code - after all you're compiling the body at that point anyway - there's no real difference between code in the arguments and code in the body.
    – Brian
    Commented Jul 16, 2009 at 9:39
  • 11
    True, but it would still slow Python down, and it would actually be quite surprising, unless you do the same for class definitions, which would make it stupidly slow as you would have to re-run the whole class definition each time you instantiate a class. As mentioned, the fix would be more surprising than the problem. Commented Jul 16, 2009 at 11:49
  • Agreed with Lennart. As Guido is fond of saying, for every language feature or standard library, there's someone out there using it. Commented Jul 16, 2009 at 13:21
  • 9
    Changing it now would be insanity--we're just exploring why it is the way it is. If it did late default evaluation to begin with, it wouldn't necessarily be surprising. It's definitely true that such a core a difference of parsing would have sweeping, and probably many obscure, effects on the language as a whole. Commented Jul 16, 2009 at 18:10
35

This actually has nothing to do with default values, other than that it often comes up as an unexpected behaviour when you write functions with mutable default values.

>>> def foo(a):
    a.append(5)
    print a

>>> a  = [5]
>>> foo(a)
[5, 5]
>>> foo(a)
[5, 5, 5]
>>> foo(a)
[5, 5, 5, 5]
>>> foo(a)
[5, 5, 5, 5, 5]

No default values in sight in this code, but you get exactly the same problem.

The problem is that foo is modifying a mutable variable passed in from the caller, when the caller doesn't expect this. Code like this would be fine if the function was called something like append_5; then the caller would be calling the function in order to modify the value they pass in, and the behaviour would be expected. But such a function would be very unlikely to take a default argument, and probably wouldn't return the list (since the caller already has a reference to that list; the one it just passed in).

Your original foo, with a default argument, shouldn't be modifying a whether it was explicitly passed in or got the default value. Your code should leave mutable arguments alone unless it is clear from the context/name/documentation that the arguments are supposed to be modified. Using mutable values passed in as arguments as local temporaries is an extremely bad idea, whether we're in Python or not and whether there are default arguments involved or not.

If you need to destructively manipulate a local temporary in the course of computing something, and you need to start your manipulation from an argument value, you need to make a copy.

11
  • 10
    Although related, I think this is distinct behaviour (as we expect append to change a "in-place"). That a default mutable is not re-instantiated on each call is the "unexpected" bit... at least for me. :) Commented Aug 24, 2012 at 12:27
  • 3
    @AndyHayden if the function is expected to modify the argument, why would it make sense to have a default? Commented Oct 17, 2017 at 13:31
  • 1
    @AndyHayden I left my own answer here with an expansion of that sentiment. Let me know what you think. I might add your example of cache={} into it for completeness. Commented Oct 17, 2017 at 18:02
  • 2
    @AndyHayden The point of my answer is that if you are ever astonished by accidentally mutating the default value of an argument, then you have another bug, which is that your code can accidentally mutate a caller's value when the default wasn't used. And note that using None and assigning the real default if the arg is None does not resolve that problem (I consider it an anti pattern for that reason). If you fix the other bug by avoiding mutating argument values whether or not they have defaults then you'll never notice or care about this "astonishing" behavior.
    – Ben
    Commented Oct 17, 2017 at 21:44
  • 1
    @AndyHayden That's the subtle thing though, what happens in the case you describe if the caller of the constructor provides a value instead of using the default? Now you've gone and aliased your object's internal attribute to an external value owned by the caller! That sort of thing is a very rich source of hard-to-track-down bugs; it's almost worse than accidentally storing the default value and then mutating it (since that at least should show head-scratching behaviour pretty soon, if you instantiate the class more than once).
    – Ben
    Commented Oct 18, 2017 at 0:30
35

Python: The Mutable Default Argument

Default arguments get evaluated at the time the function is compiled into a function object, at the start of the program runtime. When used by the function, multiple times by that function, they are and remain the same object in memory, and when mutated (if the object is of a mutable type) they remain mutated on consecutive calls.

They are mutated and stay mutated because they are the same object each time the function is called.

Equivalent code:

Since the list is bound to the function when the function object is compiled and instantiated, this:

def foo(mutable_default_argument=[]): # make a list the default argument
    """function that uses a list"""

is almost exactly equivalent to this:

_a_list = [] # create a list in the globals

def foo(mutable_default_argument=_a_list): # make it the default argument
    """function that uses a list"""

del _a_list # remove globals name binding

Demonstration

Here's a demonstration - you can verify that they are the same object each time they are referenced by

  • seeing that the list is created before the function has finished compiling to a function object,
  • observing that the id is the same each time the list is referenced,
  • observing that the list stays changed when the function that uses it is called a second time,
  • observing the order in which the output is printed from the source (which I conveniently numbered for you):

example.py

print('1. Global scope being evaluated')

def create_list():
    '''noisily create a list for usage as a kwarg'''
    l = []
    print('3. list being created and returned, id: ' + str(id(l)))
    return l

print('2. example_function about to be compiled to an object')

def example_function(default_kwarg1=create_list()):
    print('appending "a" in default default_kwarg1')
    default_kwarg1.append("a")
    print('list with id: ' + str(id(default_kwarg1)) + 
          ' - is now: ' + repr(default_kwarg1))

print('4. example_function compiled: ' + repr(example_function))


if __name__ == '__main__':
    print('5. calling example_function twice!:')
    example_function()
    example_function()

and running it with python example.py:

1. Global scope being evaluated
2. example_function about to be compiled to an object
3. list being created and returned, id: 140502758808032
4. example_function compiled: <function example_function at 0x7fc9590905f0>
5. calling example_function twice!:
appending "a" in default default_kwarg1
list with id: 140502758808032 - is now: ['a']
appending "a" in default default_kwarg1
list with id: 140502758808032 - is now: ['a', 'a']

Does this violate the principle of "Least Astonishment"?

This order of execution is frequently confusing to new users of Python. If you understand the Python execution model, then it becomes quite expected.

The usual instruction to new Python users:

But this is why the usual instruction to new users is to create their default arguments like this instead:

def example_function_2(default_kwarg=None):
    if default_kwarg is None:
        default_kwarg = []

This uses the None singleton as a sentinel object to tell the function whether or not we've gotten an argument other than the default. If we get no argument, then we actually want to use a new empty list, [], as the default.

As the tutorial section on control flow says:

If you don’t want the default to be shared between subsequent calls, you can write the function like this instead:

def f(a, L=None):
    if L is None:
        L = []
    L.append(a)
    return L
30

Already busy topic, but from what I read here, the following helped me realizing how it's working internally:

def bar(a=[]):
     print id(a)
     a = a + [1]
     print id(a)
     return a

>>> bar()
4484370232
4484524224
[1]
>>> bar()
4484370232
4484524152
[1]
>>> bar()
4484370232 # Never change, this is 'class property' of the function
4484523720 # Always a new object 
[1]
>>> id(bar.func_defaults[0])
4484370232
1
  • 4
    actually this might be a bit confusing for newcomers as a = a + [1] overloads a... consider changing it to b = a + [1] ; print id(b) and add a line a.append(2). That will make it more obvious that + on two lists always creates a new list (assigned to b), while a modified a can still have the same id(a).
    – Jörn Hees
    Commented Apr 8, 2017 at 13:47
30

The shortest answer would probably be "definition is execution", therefore the whole argument makes no strict sense. As a more contrived example, you may cite this:

def a(): return []

def b(x=a()):
    print x

Hopefully it's enough to show that not executing the default argument expressions at the execution time of the def statement isn't easy or doesn't make sense, or both.

I agree it's a gotcha when you try to use default constructors, though.

28

It's a performance optimization. As a result of this functionality, which of these two function calls do you think is faster?

def print_tuple(some_tuple=(1,2,3)):
    print some_tuple

print_tuple()        #1
print_tuple((1,2,3)) #2

I'll give you a hint. Here's the disassembly (see http://docs.python.org/library/dis.html):

#1

0 LOAD_GLOBAL              0 (print_tuple)
3 CALL_FUNCTION            0
6 POP_TOP
7 LOAD_CONST               0 (None)
10 RETURN_VALUE

#2

 0 LOAD_GLOBAL              0 (print_tuple)
 3 LOAD_CONST               4 ((1, 2, 3))
 6 CALL_FUNCTION            1
 9 POP_TOP
10 LOAD_CONST               0 (None)
13 RETURN_VALUE

I doubt the experienced behavior has a practical use (who really used static variables in C, without breeding bugs ?)

As you can see, there is a performance benefit when using immutable default arguments. This can make a difference if it's a frequently called function or the default argument takes a long time to construct. Also, bear in mind that Python isn't C. In C you have constants that are pretty much free. In Python you don't have this benefit.

0
25

This behavior is not surprising if you take the following into consideration:

  1. The behavior of read-only class attributes upon assignment attempts, and that
  2. Functions are objects (explained well in the accepted answer).

The role of (2) has been covered extensively in this thread. (1) is likely the astonishment causing factor, as this behavior is not "intuitive" when coming from other languages.

(1) is described in the Python tutorial on classes. In an attempt to assign a value to a read-only class attribute:

...all variables found outside of the innermost scope are read-only (an attempt to write to such a variable will simply create a new local variable in the innermost scope, leaving the identically named outer variable unchanged).

Look back to the original example and consider the above points:

def foo(a=[]):
    a.append(5)
    return a

Here foo is an object and a is an attribute of foo (available at foo.func_defs[0]). Since a is a list, a is mutable and is thus a read-write attribute of foo. It is initialized to the empty list as specified by the signature when the function is instantiated, and is available for reading and writing as long as the function object exists.

Calling foo without overriding a default uses that default's value from foo.func_defs. In this case, foo.func_defs[0] is used for a within function object's code scope. Changes to a change foo.func_defs[0], which is part of the foo object and persists between execution of the code in foo.

Now, compare this to the example from the documentation on emulating the default argument behavior of other languages, such that the function signature defaults are used every time the function is executed:

def foo(a, L=None):
    if L is None:
        L = []
    L.append(a)
    return L

Taking (1) and (2) into account, one can see why this accomplishes the desired behavior:

  • When the foo function object is instantiated, foo.func_defs[0] is set to None, an immutable object.
  • When the function is executed with defaults (with no parameter specified for L in the function call), foo.func_defs[0] (None) is available in the local scope as L.
  • Upon L = [], the assignment cannot succeed at foo.func_defs[0], because that attribute is read-only.
  • Per (1), a new local variable also named L is created in the local scope and used for the remainder of the function call. foo.func_defs[0] thus remains unchanged for future invocations of foo.
22

It may be true that:

  1. Someone is using every language/library feature, and
  2. Switching the behavior here would be ill-advised, but

it is entirely consistent to hold to both of the features above and still make another point:

  1. It is a confusing feature and it is unfortunate in Python.

The other answers, or at least some of them either make points 1 and 2 but not 3, or make point 3 and downplay points 1 and 2. But all three are true.

It may be true that switching horses in midstream here would be asking for significant breakage, and that there could be more problems created by changing Python to intuitively handle Stefano's opening snippet. And it may be true that someone who knew Python internals well could explain a minefield of consequences. However,

The existing behavior is not Pythonic, and Python is successful because very little about the language violates the principle of least astonishment anywhere near this badly. It is a real problem, whether or not it would be wise to uproot it. It is a design flaw. If you understand the language much better by trying to trace out the behavior, I can say that C++ does all of this and more; you learn a lot by navigating, for instance, subtle pointer errors. But this is not Pythonic: people who care about Python enough to persevere in the face of this behavior are people who are drawn to the language because Python has far fewer surprises than other language. Dabblers and the curious become Pythonistas when they are astonished at how little time it takes to get something working--not because of a design fl--I mean, hidden logic puzzle--that cuts against the intuitions of programmers who are drawn to Python because it Just Works.

11
  • 7
    -1 Although a defensible perspective, this not an answer, and I disagree with it. Too many special exceptions beget their own corner cases.
    – Marcin
    Commented Jul 7, 2012 at 19:24
  • 6
    So then, it is "amazingly ignorant" to say that in Python it would make more sense for a default argument of [] to remain [] every time the function is called? Commented Dec 27, 2012 at 22:09
  • 4
    And it is ignorant to consider as an unfortunate idiom setting a default argument to None, and then in the body of the body of the function setting if argument == None: argument = []? Is it ignorant to consider this idiom unfortunate as often people want what a naive newcomer would expect, that if you assign f(argument = []), argument will automatically default to a value of []? Commented Dec 27, 2012 at 22:11
  • 5
    But in Python, part of the spirit of the language is that you don't have to take too many deep dives; array.sort() works, and works regardless of how little you understand about sorting, big-O, and constants. The beauty of Python in the array sorting mechanism, to give one of innumerable examples, is that you are not required to take a deep dive into internals. And to say it differently, the beauty of Python is that one is not ordinarily required to take a deep dive into implementation to get something that Just Works. And there is a workaround (...if argument == None: argument = []), FAIL. Commented Dec 27, 2012 at 22:41
  • 4
    As a standalone, the statement x=[] means "create an empty list object, and bind the name 'x' to it." So, in def f(x=[]), an empty list is also created. It doesn't always get bound to x, so instead it gets bound to the default surrogate. Later when f() is called, the default is hauled out and bound to x. Since it was the empty list itself that was squirreled away, that same list is the only thing available to bind to x, whether anything has been stuck inside it or not. How could it be otherwise?
    – Jerry B
    Commented Oct 5, 2013 at 6:18
22

A simple workaround using None

>>> def bar(b, data=None):
...     data = data or []
...     data.append(b)
...     return data
... 
>>> bar(3)
[3]
>>> bar(3)
[3]
>>> bar(3)
[3]
>>> bar(3, [34])
[34, 3]
>>> bar(3, [34])
[34, 3]
1
  • This isn't an answer to the question.
    – Flimm
    Commented Mar 30, 2022 at 8:59
20

I am going to demonstrate an alternative structure to pass a default list value to a function (it works equally well with dictionaries).

As others have extensively commented, the list parameter is bound to the function when it is defined as opposed to when it is executed. Because lists and dictionaries are mutable, any alteration to this parameter will affect other calls to this function. As a result, subsequent calls to the function will receive this shared list which may have been altered by any other calls to the function. Worse yet, two parameters are using this function's shared parameter at the same time oblivious to the changes made by the other.

Wrong Method (probably...):

def foo(list_arg=[5]):
    return list_arg

a = foo()
a.append(6)
>>> a
[5, 6]

b = foo()
b.append(7)
# The value of 6 appended to variable 'a' is now part of the list held by 'b'.
>>> b
[5, 6, 7]  

# Although 'a' is expecting to receive 6 (the last element it appended to the list),
# it actually receives the last element appended to the shared list.
# It thus receives the value 7 previously appended by 'b'.
>>> a.pop()             
7

You can verify that they are one and the same object by using id:

>>> id(a)
5347866528

>>> id(b)
5347866528

Per Brett Slatkin's "Effective Python: 59 Specific Ways to Write Better Python", Item 20: Use None and Docstrings to specify dynamic default arguments (p. 48)

The convention for achieving the desired result in Python is to provide a default value of None and to document the actual behaviour in the docstring.

This implementation ensures that each call to the function either receives the default list or else the list passed to the function.

Preferred Method:

def foo(list_arg=None):
   """
   :param list_arg:  A list of input values. 
                     If none provided, used a list with a default value of 5.
   """
   if not list_arg:
       list_arg = [5]
   return list_arg

a = foo()
a.append(6)
>>> a
[5, 6]

b = foo()
b.append(7)
>>> b
[5, 7]

c = foo([10])
c.append(11)
>>> c
[10, 11]

There may be legitimate use cases for the 'Wrong Method' whereby the programmer intended the default list parameter to be shared, but this is more likely the exception than the rule.

0
17

The solutions here are:

  1. Use None as your default value (or a nonce object), and switch on that to create your values at runtime; or
  2. Use a lambda as your default parameter, and call it within a try block to get the default value (this is the sort of thing that lambda abstraction is for).

The second option is nice because users of the function can pass in a callable, which may be already existing (such as a type)

1
  • 1
    This doesn't answer the question.
    – Flimm
    Commented Mar 30, 2022 at 9:00
16

You can get round this by replacing the object (and therefore the tie with the scope):

def foo(a=[]):
    a = list(a)
    a.append(5)
    return a

Ugly, but it works.

4
  • 3
    This is a nice solution in cases where you're using automatic documentation generation software to document the types of arguments expected by the function. Putting a=None and then setting a to [] if a is None doesn't help a reader understand at a glance what is expected. Commented Jan 20, 2013 at 6:55
  • Cool idea: rebinding that name guarantees it can never be modified. I really like that.
    – holdenweb
    Commented Jan 16, 2018 at 14:29
  • This is exactly the way to do it. Python doesn't make a copy of the parameter, so it's up to you to make the copy explicitly. Once you have a copy, it's yours to modify as you please without any unexpected side effects. Commented May 25, 2018 at 16:56
  • This doesn't answer the question, though.
    – Flimm
    Commented Mar 30, 2022 at 9:00
16

When we do this:

def foo(a=[]):
    ...

... we assign the argument a to an unnamed list, if the caller does not pass the value of a.

To make things simpler for this discussion, let's temporarily give the unnamed list a name. How about pavlo ?

def foo(a=pavlo):
   ...

At any time, if the caller doesn't tell us what a is, we reuse pavlo.

If pavlo is mutable (modifiable), and foo ends up modifying it, an effect we notice the next time foo is called without specifying a.

So this is what you see (Remember, pavlo is initialized to []):

 >>> foo()
 [5]

Now, pavlo is [5].

Calling foo() again modifies pavlo again:

>>> foo()
[5, 5]

Specifying a when calling foo() ensures pavlo is not touched.

>>> ivan = [1, 2, 3, 4]
>>> foo(a=ivan)
[1, 2, 3, 4, 5]
>>> ivan
[1, 2, 3, 4, 5]

So, pavlo is still [5, 5].

>>> foo()
[5, 5, 5]
16

I sometimes exploit this behavior as an alternative to the following pattern:

singleton = None

def use_singleton():
    global singleton

    if singleton is None:
        singleton = _make_singleton()

    return singleton.use_me()

If singleton is only used by use_singleton, I like the following pattern as a replacement:

# _make_singleton() is called only once when the def is executed
def use_singleton(singleton=_make_singleton()):
    return singleton.use_me()

I've used this for instantiating client classes that access external resources, and also for creating dicts or lists for memoization.

Since I don't think this pattern is well known, I do put a short comment in to guard against future misunderstandings.

2
  • 2
    I prefer to add a decorator for memoization, and put the memoization cache onto the function object itself. Commented Feb 6, 2015 at 7:34
  • This example doesn't replace the more complex pattern you show, because you call _make_singleton at def time in the default argument example, but at call time in the global example. A true substitution would use some sort of mutable box for the default argument value, but the addition of the argument makes an opportunity to pass alternate values. Commented Nov 19, 2017 at 8:29
15

Every other answer explains why this is actually a nice and desired behavior, or why you shouldn't be needing this anyway. Mine is for those stubborn ones who want to exercise their right to bend the language to their will, not the other way around.

We will "fix" this behavior with a decorator that will copy the default value instead of reusing the same instance for each positional argument left at its default value.

import inspect
from copy import deepcopy  # copy would fail on deep arguments like nested dicts

def sanify(function):
    def wrapper(*a, **kw):
        # store the default values
        defaults = inspect.getargspec(function).defaults # for python2
        # construct a new argument list
        new_args = []
        for i, arg in enumerate(defaults):
            # allow passing positional arguments
            if i in range(len(a)):
                new_args.append(a[i])
            else:
                # copy the value
                new_args.append(deepcopy(arg))
        return function(*new_args, **kw)
    return wrapper

Now let's redefine our function using this decorator:

@sanify
def foo(a=[]):
    a.append(5)
    return a

foo() # '[5]'
foo() # '[5]' -- as desired

This is particularly neat for functions that take multiple arguments. Compare:

# the 'correct' approach
def bar(a=None, b=None, c=None):
    if a is None:
        a = []
    if b is None:
        b = []
    if c is None:
        c = []
    # finally do the actual work

with

# the nasty decorator hack
@sanify
def bar(a=[], b=[], c=[]):
    # wow, works right out of the box!

It's important to note that the above solution breaks if you try to use keyword args, like so:

foo(a=[4])

The decorator could be adjusted to allow for that, but we leave this as an exercise for the reader ;)

2
  • 1
    This also breaks if the default argument is deep, like {"grandparent": {"parent": {"child": "value"}}} . Only the top-level dictionary gets copied by value, the other dictionaries get copied by reference. This problem occurs because you used copy instead of deepcopy
    – Flimm
    Commented Mar 30, 2022 at 8:52
  • 1
    @Flimm I find your phrase "this breaks" rather unfair as it seems to suggests the entire concept is somehow flawed, while it is in fact only a minor detail of the implementation. But thank you for the comment nonetheless, I shall edit and improve my answer.
    – Przemek D
    Commented Mar 31, 2022 at 10:07
14

Yes, this is a design flaw in Python

I've read all the other answers and I'm not convinced. This design does violate the principle of least astonishment.

The defaults could have been designed to be evaluated when the function is called, rather than when the function is defined. This is how Javascript does it:

function foo(a=[]) {
  a.push(5);
  return a;
}
console.log(foo()); // [5]
console.log(foo()); // [5]
console.log(foo()); // [5]

As further evidence that this is a design flaw, Python core developers are currently discussing introducing new syntax to fix this problem. See this article: Late-bound argument defaults for Python.

For even more evidence that this a design flaw, if you Google "Python gotchas", this design is mentioned as a gotcha, usually the first gotcha in the list, in the first 9 Google results (1, 2, 3, 4, 5, 6, 7, 8, 9). In contrast, if you Google "Javascript gotchas", the behaviour of default arguments in Javascript is not mentioned as a gotcha even once.

Gotchas, by definition, violate the principle of least astonishment. They astonish. Given there are superiour designs for the behaviour of default argument values, the inescapable conclusion is that Python's behaviour here represents a design flaw.

I say this as someone who loves Python. We can be fans of Python, and still admit that everyone who is unpleasantly surprised by this aspect of Python is unpleasantly surprised because it is a genuine "gotcha".

6
  • 1
    Calling it a design flaw assumes there is a better solution. But that solution (whichever one is chosen in particular) would break the orthogonality of Python (from which the incriminated behavior follows), which would create lots more surprise in other places. Commented Apr 26, 2023 at 8:47
  • 1
    @LutzPrechelt There is a better design, namely, the one used in some other programming languages, like JavaScript, as I mentioned in my answer. I don't understand how the better design would break things in Python, except for backwards compatibility. What do you mean by "orthogonality"?
    – Flimm
    Commented Apr 26, 2023 at 9:12
  • 1
    by orthogonality I mean that in Python def is a statement that is executed at module load time (for a top-level function). The default argument definition is part of that statement and therefore ought to be executed at load time as well -- and any other behavior would also be surprising. Commented Apr 27, 2023 at 11:51
  • 2
    @LutzPrechelt Just because the def statement is executed at module load time, that does not mean that the default arg values should be executed at module load time too. Take the statement foobar = lambda x: print("hi") . Would you be surprised if hi got printed at module load time? I know I would be, even though this statement is executed at module load time. Intuitively, we expect that the print part of this statement would only be executed once the lambda is called. Likewise, as evidenced by people's well-documented surprised, we expect the default value to be executed when called
    – Flimm
    Commented Apr 27, 2023 at 14:54
  • 1
    en.wikipedia.org/wiki/Orthogonality_(programming) is a mostly good discussion of the idea. Commented Apr 28, 2023 at 21:19
9

This "bug" gave me a lot of overtime work hours! But I'm beginning to see a potential use of it (but I would have liked it to be at the execution time, still)

I'm gonna give you what I see as a useful example.

def example(errors=[]):
    # statements
    # Something went wrong
    mistake = True
    if mistake:
        tryToFixIt(errors)
        # Didn't work.. let's try again
        tryToFixItAnotherway(errors)
        # This time it worked
    return errors

def tryToFixIt(err):
    err.append('Attempt to fix it')

def tryToFixItAnotherway(err):
    err.append('Attempt to fix it by another way')

def main():
    for item in range(2):
        errors = example()
    print '\n'.join(errors)

main()

prints the following

Attempt to fix it
Attempt to fix it by another way
Attempt to fix it
Attempt to fix it by another way
1
  • 2
    Your example doesn't seem very realistic. Why would you pass errors as a parameter rather than starting from scratch every time? Commented May 25, 2021 at 23:39
8

This is not a design flaw. Anyone who trips over this is doing something wrong.

There are 3 cases I see where you might run into this problem:

  1. You intend to modify the argument as a side effect of the function. In this case it never makes sense to have a default argument. The only exception is when you're abusing the argument list to have function attributes, e.g. cache={}, and you wouldn't be expected to call the function with an actual argument at all.
  2. You intend to leave the argument unmodified, but you accidentally did modify it. That's a bug, fix it.
  3. You intend to modify the argument for use inside the function, but didn't expect the modification to be viewable outside of the function. In that case you need to make a copy of the argument, whether it was the default or not! Python is not a call-by-value language so it doesn't make the copy for you, you need to be explicit about it.

The example in the question could fall into category 1 or 3. It's odd that it both modifies the passed list and returns it; you should pick one or the other.

10
  • 1
    "Doing something wrong" is the diagnosis. That said, I think there are times were =None pattern is useful, but generally you don't want to modify if passed a mutable in that case (2). The cache={} pattern is really an interview-only solution, in real code you probably want @lru_cache! Commented Oct 17, 2017 at 18:19
  • 5
    Totally disagree, its absolutely a design flaw in many cases and not the programmer doing something wong
    – aCuria
    Commented May 25, 2021 at 10:44
  • 1
    I have never run into the problem of the OP even though it is so highly upvoted, because having a default argument be mutable is weird design to me.
    – qwr
    Commented May 25, 2021 at 19:56
  • 2
    @MarkRansom If we take it as given that side effects are OK, there's nothing wrong with modifying a default argument as part of a side-effect-ful function. Let's say you have a function that does something to a list and returns the list. We want to ensure that the function always returns a list. Then having an empty (or non-empty) list as a default makes perfect sense. The language is violating a large proportion of new Python programmers' expectations. Why are they wrong and the language right? Would you be making the opposite argument if the language had the opposite behavior? Commented Jun 3, 2021 at 14:05
  • 2
    @MarkRansom No, they're not; for example, JavaScript doesn't have this design flaw. Commented Jun 7, 2021 at 15:30
8

Just change the function to be:

def notastonishinganymore(a = []): 
    '''The name is just a joke :)'''
    a = a[:]
    a.append(5)
    return a
1
  • This doesn't answer the question, though.
    – Flimm
    Commented Mar 30, 2022 at 9:01
8

TLDR: Define-time defaults are consistent and strictly more expressive.


Defining a function affects two scopes: the defining scope containing the function, and the execution scope contained by the function. While it is pretty clear how blocks map to scopes, the question is where def <name>(<args=defaults>): belongs to:

...                           # defining scope
def name(parameter=default):  # ???
    ...                       # execution scope

The def name part must evaluate in the defining scope - we want name to be available there, after all. Evaluating the function only inside itself would make it inaccessible.

Since parameter is a constant name, we can "evaluate" it at the same time as def name. This also has the advantage it produces the function with a known signature as name(parameter=...):, instead of a bare name(...):.

Now, when to evaluate default?

Consistency already says "at definition": everything else of def <name>(<args=defaults>): is best evaluated at definition as well. Delaying parts of it would be the astonishing choice.

The two choices are not equivalent, either: If default is evaluated at definition time, it can still affect execution time. If default is evaluated at execution time, it cannot affect definition time. Choosing "at definition" allows expressing both cases, while choosing "at execution" can express only one:

def name(parameter=defined):  # set default at definition time
    ...

def name(parameter=default):     # delay default until execution time
    parameter = default if parameter is None else parameter
    ...
6
  • 2
    "Consistency already says "at definition": everything else of def <name>(<args=defaults>): is best evaluated at definition as well." I don't think the conclusion follows from the premise. Just because two things are on the same line doesn't mean they should be evaluated in the same scope. default is a different thing than the rest of the line: it's an expression. Evaluating an expression is a very different process from defining a function.
    – LarsH
    Commented Sep 23, 2019 at 14:38
  • 1
    @LarsH Function definitions are are evaluated in Python. Whether that is from a statement (def) or expression (lambda) does not change that creating a function means evaluation -- especially of its signature. And defaults are part of a function's signature. That does not mean defaults have to be evaluated immediately -- type hints may not, for example. But it certainly suggests they should unless there is a good reason not to. Commented Sep 23, 2019 at 15:22
  • 2
    OK, creating a function means evaluation in some sense, but obviously not in the sense that every expression within it is evaluated at the time of definition. Most aren't. It's not clear to me in what sense the signature is especially "evaluated" at definition time any more than the function body is "evaluated" (parsed into a suitable representation); whereas expressions in the function body are clearly not evaluated in the full sense. From this point of view, consistency would say that expressions in the signature shouldn't be "fully" evaluated either.
    – LarsH
    Commented Sep 24, 2019 at 15:09
  • 2
    I don't mean that you're wrong, only that your conclusion doesn't follow from consistency alone.
    – LarsH
    Commented Sep 24, 2019 at 15:09
  • @LarsH Defaults are neither part of the body, nor am I claiming that consistency is the only criteria. Can you make a suggestion how to clarify the answer? Commented Sep 24, 2019 at 15:15

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