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 to always return a list with only one element:
. The result is instead very different, and very astonishing (for a novice):
>>> foo()  >>> foo() [5, 5] >>> foo() [5, 5, 5] >>> foo() [5, 5, 5, 5] >>> foo()
A manager of mine once had his first encountered 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 ?)
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()  >>> 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.
Guido is a fantastic designer.
I reread all the very interesting and good answers you provided, and it was hard to assign a "correct tickmark", as everyone had good points in the answer. I marked Roberto's answer as correct because it was simpler and revealing, so that newcomers browsing this question can start from his answer and then delve into remaining more complex (but very insightful) answers.
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 get to think into 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, Effbot 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.
Suppose you have the following code
fruits = ("apples", "bannanas", "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", "bannanas", "loganberries")
However, supposed 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 then 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 ovject 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 builtin 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.
AFAICS no one has yet posted the relevant part of the documentation :
Default parameter values are evaluated 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 [...]
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
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:
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):
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.
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 classes. 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.
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:
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
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)]
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.
This behavior is easy explained by:
def x(a=0, b=, c=, d=0): a = a + 1 b = b +  c.append(1) print a, b, c
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.
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):
0 LOAD_GLOBAL 0 (print_tuple) 3 CALL_FUNCTION 0 6 POP_TOP 7 LOAD_CONST 0 (None) 10 RETURN_VALUE
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.
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 =  >>> 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).
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.
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.
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
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 + "
Many builtin mutable types have a copy method like
some_set.copy() or can be copied easy like
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.
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() print somevar # [1, 2, ] but usually expected [1, 2] print t2._var1 # [1, 2, ] 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 )
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.) .)
Only if the side effect on the actual parameter is required but unwanted on the default parameter then the useful solution is
if var1 is None:
var1 = 
3) In some cases is the mutable behavior of default parameters useful . http://effbot.org/pyfaq/how-do-i-copy-an-object-in-python.htm
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.
This behavior is not surprising if you take the following into consideration:
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
foo is an object and
a is an attribute of
foo (available at
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.
foo without overriding a default uses that default's value from
foo.func_defs. In this case,
foo.func_defs is used for
a within function object's code scope. Changes to
foo.func_defs, which is part of the
foo object and persists between execution of the code in
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 the desired behavior:
foofunction object is instantiated,
foo.func_defsis set to
None, an immutable object.
Lin the function call),
None) is available in the local scope as
L = , the assignment cannot succeed at
foo.func_defs, because that attribute is read-only.
Lis created in the local scope and used for the remainder of the function call.
foo.func_defsthus remains unchanged for future invocations of
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.
A simple workaround using None
>>> def bar(b, data=None): ... data = data or  ... data.append(b) ... return data ... >>> bar(3)  >>> bar(3)  >>> bar(3)  >>> bar(3, ) [34, 3] >>> bar(3, ) [34, 3]
The solutions here are:
Noneas your default value (or a nonce
object), and switch on that to create your values at runtime; or
lambdaas 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
It may be true that:
it is entirely consistent to hold to both of the features above and still make another point:
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.
>>> def a(): >>> print "a executed" >>> return  >>> x =a() a executed >>> def b(m=): >>> m.append(5) >>> print m >>> b(x)  >>> b(x) [5, 5]
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
I think the answer to this question lies in how python pass data to parameter (pass by value or by reference), not mutability or how python handle the "def" statement.
A brief introduction. First, there are two type of data types in python, one is simple elementary data type, like numbers, and another data type is objects. Second, when passing data to parameters, python pass elementary data type by value, i.e., make a local copy of the value to a local variable, but pass object by reference, i.e., pointers to the object.
Admitting the above two points, let's explain what happened to the python code. It's only because of passing by reference for objects, but has nothing to do with mutable/immutable, or arguably the fact that "def" statement is executed only once when it is defined.
 is an object, so python pass the reference of  to
a is only a pointer to  which lies in memory as an object. There is only one copy of  with, however, many references to it. For the first foo(), the list  is changed to
 by append method. But Note that there is only one copy of the list object and this object now becomes
. When running the second foo(), what effbot webpage says (items is not evaluated any more) is wrong.
a is evaluated to be the list object, although now the content of the object is
. This is the effect of passing by reference! The result of foo(3) can be easily derived in the same way.
To further validate my answer, let's take a look at two additional codes.
====== No. 2 ========
def foo(x, items=None): if items is None: items =  items.append(x) return items foo(1) #return  foo(2) #return  foo(3) #return 
 is an object, so is
None (the former is mutable while the latter is immutable. But the mutability has nothing to do with the question). None is somewhere in the space but we know it's there and there is only one copy of None there. So every time foo is invoked, items is evaluated (as opposed to some answer that it is only evaluated once) to be None, to be clear, the reference (or the address) of None. Then in the foo, item is changed to , i.e., points to another object which has a different address.
====== No. 3 =======
def foo(x, items=): items.append(x) return items foo(1) # returns  foo(2,) # returns  foo(3) # returns [1,3]
The invocation of foo(1) make items point to a list object  with an address, say, 11111111. the content of the list is changed to
 in the foo function in the sequel, but the address is not changed, still 11111111. Then foo(2,) is coming. Although the  in foo(2,) has the same content as the default parameter  when calling foo(1), their address are different! Since we provide the parameter explicitly,
items has to take the address of this new
, say 2222222, and return it after making some change. Now foo(3) is executed. since only
x is provided, items has to take its default value again. What's the default value? It is set when defining the foo function: the list object located in 11111111. So the items is evaluated to be the address 11111111 having an element 1. The list located at 2222222 also contains one element 2, but it is not pointed by items any more. Consequently, An append of 3 will make
From the above explanations, we can see that the effbot  webpage recommended in the accepted answer failed to give a relevant answer to this question. What is more, I think a point in the effbot webpage is wrong. I think the code regarding the UI.Button is correct:
for i in range(10): def callback(): print "clicked button", i UI.Button("button %s" % i, callback)
Each button can hold a distinct callback function which will display different value of
i. I can provide an example to show this:
x= for i in range(10): def callback(): print(i) x.append(callback)
If we execute
x() we'll get 7 as expected, and
x() will gives 9, another value of
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
def foo(a=pavlo): ...
At any time, if the caller doesn't tell us what
a is, we reuse
pavlo is mutable (modifiable), and
foo ends up modifying it, an effect we notice the next time
foo is called without specifying
So this is what you see (Remember,
pavlo is initialized to ):
>>> foo() 
pavlo is .
foo() again modifies
>>> foo() [5, 5]
a when calling
pavlo is not touched.
>>> ivan = [1, 2, 3, 4] >>> foo(a=ivan) [1, 2, 3, 4, 5] >>> ivan [1, 2, 3, 4, 5]
pavlo is still
>>> foo() [5, 5, 5]