Related
i = 5
def f(arg=i):
print(arg)
i = 6
f()
I am learning Python from the official documentation. There I find the above piece of code which I am unable to understand as to why 5 is printed instead of 6. I am relatively new to Python. Can somebody help me understand the concept?
def f(arg=i) says "make me a function f where the default value for arg is whatever i is right now". At the time of defining the function, i=5.
i = 5
def f(arg=i)
print(arg)
The i is evaluated at the time of definition, so the code above has the same meaning as the code below:
def f(arg=5)
print(arg)
This means that, when the function is called without arguments, arg will have the value 5, no matter what the value of i is now.
In order to get what you want, just do the following:
def f(arg)
print(arg)
i = 6
f(i)
Because the function takes its default value on the first declaration of 'i'.
Change to i=6 on the first line if you want you code to print 6.
Hope I helped !
This is the difference between something being handled by reference vs by value. When you defined the function f you told it to set the argument's default value to i this is done by value, not by reference, so it took whatever the value of i was at that time and set the default for the function to that. Changing the value of i after that point does not change the value of arg. If you want it to work that way you could do this:
i = 5
def f(arg = None):
if (arg = None)
arg = i
print(arg)
i = 6
f()
This lets you pass a value for arg into the function as normal, but if you don't (or you explicitly pass None) it updates arg to the current value of i if arg is still None (Python's version of NULL if you're familiar with other languages)
Something similar can be done using the or operator, arg = arg or i,but that will check if arg is falsy, and when using integers like you are in your example, 0 will be caught by the check.
What others have said is true...the default is evaluated at the time of function creation, but it is not that it takes the "value of i" at the time of creation. The default is assigned the object referred to by "i" at the time of creation. This is an important point, because if that object is mutable, the default can be changed!
Here's what happens:
import inspect
i = 5 # name "i" refers to an immutable Python integer object of value 5.
print(f'i = {i} (id={id(i)})') # Note value and ID
# Create function "f" with a parameter whose default is the object
# referred to by name "i" *at this point*.
def f(arg=i):
print(f'arg = {arg} (id={id(arg)})')
# Use the inspect module to extract the defaults from the function.
# Note the value and ID
defaults = dict(inspect.getmembers(f))['__defaults__']
print(f'defaults = {defaults} (id={id(defaults[0])})')
# name "i" now refers to a different immutable Python integer object of value 6.
i = 6
print(f'i = {i} (id={id(i)})') # Note value and ID (changed!)
f() # default for function still referes to object 5.
f(i) # override the default with object currently referred to by name "i"
Output:
i = 5 (id=2731452426672) # Original object
defaults = (5,) (id=2731452426672) # default refers to same object
i = 6 (id=2731452426704) # Different object
arg = 5 (id=2731452426672) # f() default is the original object
arg = 6 (id=2731452426704) # f(i) parameter is different object
Now see the results of a mutable default:
import inspect
i = [5] # name "i" refers to an mutable Python list containing immutable integer object 5
print(f'i = {i} (id={id(i)})') # Note value and ID
# Create function "f" with a parameter whose default is the object
# referred to by name "i" *at this point*.
def f(arg=i):
print(f'arg = {arg} (id={id(arg)})')
# Use the inspect module to extract the defaults from the function.
# Note the value and ID
defaults = dict(inspect.getmembers(f))['__defaults__']
print(f'defaults = {defaults} (id={id(defaults[0])})')
# name "i" now refers to a different immutable Python integer object of value 6.
i[0] = 6 # MUTATE the content of the object "i" refers to.
print(f'i = {i} (id={id(i)})') # Note value and ID (UNCHANGED!)
f() # default for function still refers to original list object, but content changed!
i = [7] # Create a different list object
print(f'i = {i} (id={id(i)})') # Note value and ID (changed)
f(i) # override the default currently refered to by name "i"
Output:
i = [5] (id=2206901216704) # Original object
defaults = ([5],) (id=2206901216704) # default refers to original object
i = [6] (id=2206901216704) # Still original object, but content changed!
arg = [6] (id=2206901216704) # f() default refers to orginal object, but content changed!
i = [7] (id=2206901199296) # Create a new list object
arg = [7] (id=2206901199296) # f(i) parameter refers to new passed object.
This can have strange side effects if not understood well:
>>> def f(a,b=[]): # mutable default
... b.append(a)
... return b
...
>>> x = f(1)
>>> x
[1]
>>> y = f(2) # some would think this would return [2]
>>> y
[1, 2]
>>> x # x changed from [1] to [1,2] as well!
[1, 2]
Above, b refers to the original default list object. Appending to it mutates the default list. Returning it makes x refer to the same object. The default list now contains [1] so appending in the 2nd call make it [1,2]. y refers to the same default object as x so both names refer see the same object content.
To fix, make the default immutable and create a new list when the default is seen:
>>> def f(a,b=None):
... if b is None:
... b = []
... b.append(a)
... return b
...
>>> f(1)
[1]
>>> f(2)
[2]
This is because you are assigning the value when the function is created. arg at the time of creation will be defaulted to what i is in that moment. Since at the time of the function being created the value of i is 5 then that's what the default value of that argument becomes. After the initial creation of the function i in the function argument is no longer linked to i in the body.
Python Tutorial 4.7.1. Default Argument Values states the following:
Important warning: The default value is evaluated only once. This makes a difference
when the default is a mutable object such as a list, dictionary, or instances of most
classes. For example, the following function accumulates the arguments passed to it on
subsequent calls:
def f(a, L=[]):
L.append(a)
return L
print f(1)
print f(2)
print f(3)
This will print
[1]
[1, 2]
[1, 2, 3]
I don't quite understand the meaning of "evaluated only once" in terms of memory management. Apparently, the default value of the function is evaluated once when the function is first called and stored in a separate memory address even after the function has ended. (according to my understanding, after the function ended, all local variables should be freed?)
Am I correct?
In Python, functions are objects too, and the defaults are stored with the function object. Defaults are not locals; it is just that when the function is called, the arguments are bound to a default when not given an explicit value.
When Python encounters a def <functionname>(<arguments>): statement, it creates a function object for you there and then; this is 'definition time'; the function is not called but merely created. It is then that defaults are evaluated and stored, in an attribute on the function object.
Then when you call the function, the defaults have already been created and are used when you didn't provide a more concrete value for the argument. Because the defaults are stored with the function object, you get to see changes to mutable objects between function calls.
The locals are still cleared up of course, but as they are references (all identifiers in Python are), the objects they were bound to are only cleared up if nothing else is referencing them anymore either.
You can take a look a the defaults of any function object:
>>> def foo(bar='spam', eggs=[]):
... eggs.append(bar)
... return eggs
...
>>> foo.__defaults__
('spam', [])
>>> foo()
['spam']
>>> foo.__defaults__
('spam', ['spam'])
>>> foo() is foo.__defaults__[1]
True
The foo() function has a __defaults__ attribute, a tuple of default values to use when no values for the arguments have been passed in. You can see the mutable list change as the function is called, and because the function returns the eggs list, you can also see that it is the exact same object as the second value in that tuple.
If you don't want your defaults to be shared and instead need a new value for a parameter every time the function is called, but the parameter is not given, you need to set the default value to a sentinel object. If your parameter is still set to that sentinel in the function body, you can execute code to set a fresh default value. None is usually the best choice:
def foo(bar='spam', eggs=None):
if eggs is None:
eggs = []
If it should be possible to use None as a non-default value, use a singleton sentinel created beforehand:
_sentinel = object()
def foo(bar='spam', eggs=_sentinel):
if eggs is _sentinel:
eggs = []
The function that you have defined f is an object in its own regard. When you define defaults, these defaults are bound to the function that you have created.
You can see this in action:
>>> def f(a, L=[]):
... L.append(a)
... return L
>>> print id(f)
4419902952
>>> print f.__defaults__
([],)
>>> f(1)
[1]
>>> print id(f)
4419902952
>>> print f.__defaults__
([1],)
edit, further, you can see that the list container does not change either:
>>> print id(f.__defaults__[0])
4419887544
>>> f(2)
[1, 2]
>>> print id(f.__defaults__[0])
4419887544
On each subsequent call, the default list ("L") of your f function will have your a value appended.
A function is just an object in python, that is created using the def syntax. Default values are stored within the function object when the function is defined, and they are not re-evaluated later.
This is sometimes used to create function variables that persist to subsequent invocations. You can use the __defaults__ methods to check what the default values are for your function.
A common way to initialize new objects instead of reusing the same is:
def f(a, L=None):
if L is None:
L = []
L.append(a)
return L
You can check this page for more details.
Sorry this answer was meant for a different question, but I'll leave it here as a reference if anyone who wants to look at it. Define once means that at the first point when the code is executed, the default variable gets assigned to an object which is retained within the function object itself.
Notice only 1 object address gets printed, the default list object is used.
def f(a, L=[]):
print("id default: ", id(L))
L.append(a)
print("id used: ", id(L)
return L
Notice 2 different object addresses are printed, when you perform L=[] within the function, you are binding L to a different list object, therefore the default list object does not get change.
def f(a, L=[]):
print("id default: ", id(L))
if L == []:
L = []
L.append(a)
print("id used: ", id(L))
return L
The function above is basically the same as the one below except it uses the None object instead of a empty list object.
def f(a, L=None):
print("id default", id(L))
if L is None:
L = []
L.append(a)
print("id used: ", id(L))
return L
This question already has answers here:
How do I pass a variable by reference?
(39 answers)
Closed 8 months ago.
In some languages you can pass a parameter by reference or value by using a special reserved word like ref or val. When you pass a parameter to a Python function it never alters the value of the parameter on leaving the function.The only way to do this is by using the global reserved word (or as i understand it currently).
Example 1:
k = 2
def foo (n):
n = n * n #clarity regarding comment below
square = n
return square
j = foo(k)
print j
print k
would show
>>4
>>2
showing k to be unchanged.
In this example the variable n is never changed
Example 2:
n = 0
def foo():
global n
n = n * n
return n
In this example the variable n is changed.
Is there any way in Python to call a function and tell Python that the parameter is either a value or reference parameter instead of using global?
There are essentially three kinds of 'function calls':
Pass by value
Pass by reference
Pass by object reference
Python is a PASS-BY-OBJECT-REFERENCE programming language.
Firstly, it is important to understand that a variable, and the value of the variable (the object) are two seperate things. The variable 'points to' the object. The variable is not the object. Again:
THE VARIABLE IS NOT THE OBJECT
Example: in the following line of code:
>>> x = []
[] is the empty list, x is a variable that points to the empty list, but x itself is not the empty list.
Consider the variable (x, in the above case) as a box, and 'the value' of the variable ([]) as the object inside the box.
PASS BY OBJECT REFERENCE (Case in python):
Here, "Object references are passed by value."
def append_one(li):
li.append(1)
x = [0]
append_one(x)
print x
Here, the statement x = [0] makes a variable x (box) that points towards the object [0].
On the function being called, a new box li is created. The contents of li are the SAME as the contents of the box x. Both the boxes contain the same object. That is, both the variables point to the same object in memory. Hence, any change to the object pointed at by li will also be reflected by the object pointed at by x.
In conclusion, the output of the above program will be:
[0, 1]
Note:
If the variable li is reassigned in the function, then li will point to a separate object in memory. x however, will continue pointing to the same object in memory it was pointing to earlier.
Example:
def append_one(li):
li = [0, 1]
x = [0]
append_one(x)
print x
The output of the program will be:
[0]
PASS BY REFERENCE:
The box from the calling function is passed on to the called function. Implicitly, the contents of the box (the value of the variable) is passed on to the called function. Hence, any change to the contents of the box in the called function will be reflected in the calling function.
PASS BY VALUE:
A new box is created in the called function, and copies of contents of the box from the calling function is stored into the new boxes.
You can not change an immutable object, like str or tuple, inside a function in Python, but you can do things like:
def foo(y):
y[0] = y[0]**2
x = [5]
foo(x)
print x[0] # prints 25
That is a weird way to go about it, however, unless you need to always square certain elements in an array.
Note that in Python, you can also return more than one value, making some of the use cases for pass by reference less important:
def foo(x, y):
return x**2, y**2
a = 2
b = 3
a, b = foo(a, b) # a == 4; b == 9
When you return values like that, they are being returned as a Tuple which is in turn unpacked.
edit:
Another way to think about this is that, while you can't explicitly pass variables by reference in Python, you can modify the properties of objects that were passed in. In my example (and others) you can modify members of the list that was passed in. You would not, however, be able to reassign the passed in variable entirely. For instance, see the following two pieces of code look like they might do something similar, but end up with different results:
def clear_a(x):
x = []
def clear_b(x):
while x: x.pop()
z = [1,2,3]
clear_a(z) # z will not be changed
clear_b(z) # z will be emptied
OK, I'll take a stab at this. Python passes by object reference, which is different from what you'd normally think of as "by reference" or "by value". Take this example:
def foo(x):
print x
bar = 'some value'
foo(bar)
So you're creating a string object with value 'some value' and "binding" it to a variable named bar. In C, that would be similar to bar being a pointer to 'some value'.
When you call foo(bar), you're not passing in bar itself. You're passing in bar's value: a pointer to 'some value'. At that point, there are two "pointers" to the same string object.
Now compare that to:
def foo(x):
x = 'another value'
print x
bar = 'some value'
foo(bar)
Here's where the difference lies. In the line:
x = 'another value'
you're not actually altering the contents of x. In fact, that's not even possible. Instead, you're creating a new string object with value 'another value'. That assignment operator? It isn't saying "overwrite the thing x is pointing at with the new value". It's saying "update x to point at the new object instead". After that line, there are two string objects: 'some value' (with bar pointing at it) and 'another value' (with x pointing at it).
This isn't clumsy. When you understand how it works, it's a beautifully elegant, efficient system.
Hope the following description sums it up well:
There are two things to consider here - variables and objects.
If you are passing a variable, then it's pass by value, which means the changes made to the variable within the function are local to that function and hence won't be reflected globally. This is more of a 'C' like behavior.
Example:
def changeval( myvar ):
myvar = 20;
print "values inside the function: ", myvar
return
myvar = 10;
changeval( myvar );
print "values outside the function: ", myvar
O/P:
values inside the function: 20
values outside the function: 10
If you are passing the variables packed inside a mutable object, like a list, then the changes made to the object are reflected globally as long as the object is not re-assigned.
Example:
def changelist( mylist ):
mylist2=['a'];
mylist.append(mylist2);
print "values inside the function: ", mylist
return
mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist
O/P:
values inside the function: [1, 2, 3, ['a']]
values outside the function: [1, 2, 3, ['a']]
Now consider the case where the object is re-assigned. In this case, the object refers to a new memory location which is local to the function in which this happens and hence not reflected globally.
Example:
def changelist( mylist ):
mylist=['a'];
print "values inside the function: ", mylist
return
mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist
O/P:
values inside the function: ['a']
values outside the function: [1, 2, 3]
Python is neither pass-by-value nor pass-by-reference. It's more of "object references are passed by value" as described here:
Here's why it's not pass-by-value. Because
def append(list):
list.append(1)
list = [0]
reassign(list)
append(list)
returns [0,1] showing that some kind of reference was clearly passed as pass-by-value does not allow a function to alter the parent scope at all.
Looks like pass-by-reference then, hu? Nope.
Here's why it's not pass-by-reference. Because
def reassign(list):
list = [0, 1]
list = [0]
reassign(list)
print list
returns [0] showing that the original reference was destroyed when list was reassigned. pass-by-reference would have returned [0,1].
For more information look here:
If you want your function to not manipulate outside scope, you need to make a copy of the input parameters that creates a new object.
from copy import copy
def append(list):
list2 = copy(list)
list2.append(1)
print list2
list = [0]
append(list)
print list
Technically python do not pass arguments by value: all by reference. But ... since python has two types of objects: immutable and mutable, here is what happens:
Immutable arguments are effectively passed by value: string, integer, tuple are all immutable object types. While they are technically "passed by reference" (like all parameters), since you can't change them in-place inside the function it looks/behaves as if it is passed by value.
Mutable arguments are effectively passed by reference: lists or dictionaries are passed by its pointers. Any in-place change inside the function like (append or del) will affect the original object.
This is how Python is designed: no copies and all are passed by reference. You can explicitly pass a copy.
def sort(array):
# do sort
return array
data = [1, 2, 3]
sort(data[:]) # here you passed a copy
Last point I would like to mention which is a function has its own scope.
def do_any_stuff_to_these_objects(a, b):
a = a * 2
del b['last_name']
number = 1 # immutable
hashmap = {'first_name' : 'john', 'last_name': 'legend'} # mutable
do_any_stuff_to_these_objects(number, hashmap)
print(number) # 1 , oh it should be 2 ! no a is changed inisde the function scope
print(hashmap) # {'first_name': 'john'}
So this is a little bit of a subtle point, because while Python only passes variables by value, every variable in Python is a reference. If you want to be able to change your values with a function call, what you need is a mutable object. For example:
l = [0]
def set_3(x):
x[0] = 3
set_3(l)
print(l[0])
In the above code, the function modifies the contents of a List object (which is mutable), and so the output is 3 instead of 0.
I write this answer only to illustrate what 'by value' means in Python. The above code is bad style, and if you really want to mutate your values you should write a class and call methods within that class, as MPX suggests.
Consider that the variable is a box and the value it points to is the "thing" inside the box:
1. Pass by reference : function shares the same box and thereby the thing inside also.
2. Pass by value : function creates a new box, a replica of the old one, including a copy of whatever thing is inside it. Eg. Java - functions create a copy of the box and the thing inside it which can be: a primitive / a reference to an object. (note that the copied reference in the new box and the original both still point to the same object, here the reference IS the thing inside the box, not the object it is pointing to)
3. Pass by object-reference: the function creates a box, but it encloses the same thing the initial box was enclosing. So in Python:
a) if the thing inside said box is mutable, changes made will reflect back in the original box (eg. lists)
b) if the thing is immutable (like python strings and numeric types), then the box inside the function will hold the same thing UNTIL you try to change its value. Once changed, the thing in the function's box is a totally new thing compared to the original one. Hence id() for that box will now give the identity of the new thing it encloses.
The answer given is
def set_4(x):
y = []
for i in x:
y.append(i)
y[0] = 4
return y
and
l = [0]
def set_3(x):
x[0] = 3
set_3(l)
print(l[0])
which is the best answer so far as it does what it says in the question. However,it does seem a very clumsy way compared to VB or Pascal.Is it the best method we have?
Not only is it clumsy, it involves mutating the original parameter in some way manually eg by changing the original parameter to a list: or copying it to another list rather than just saying: "use this parameter as a value " or "use this one as a reference". Could the simple answer be there is no reserved word for this but these are great work arounds?
class demoClass:
x = 4
y = 3
foo1 = demoClass()
foo1.x = 2
foo2 = demoClass()
foo2.y = 5
def mySquare(myObj):
myObj.x = myObj.x**2
myObj.y = myObj.y**2
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)
mySquare(foo1)
mySquare(foo2)
print('After square:')
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)
In Python the passing by reference or by value has to do with what are the actual objects you are passing.So,if you are passing a list for example,then you actually make this pass by reference,since the list is a mutable object.Thus,you are passing a pointer to the function and you can modify the object (list) in the function body.
When you are passing a string,this passing is done by value,so a new string object is being created and when the function terminates it is destroyed.
So it all has to do with mutable and immutable objects.
Python already call by ref..
let's take example:
def foo(var):
print(hex(id(var)))
x = 1 # any value
print(hex(id(x))) # I think the id() give the ref...
foo(x)
OutPut
0x50d43700 #with you might give another hex number deppend on your memory
0x50d43700
Python Tutorial 4.7.1. Default Argument Values states the following:
Important warning: The default value is evaluated only once. This makes a difference
when the default is a mutable object such as a list, dictionary, or instances of most
classes. For example, the following function accumulates the arguments passed to it on
subsequent calls:
def f(a, L=[]):
L.append(a)
return L
print f(1)
print f(2)
print f(3)
This will print
[1]
[1, 2]
[1, 2, 3]
I don't quite understand the meaning of "evaluated only once" in terms of memory management. Apparently, the default value of the function is evaluated once when the function is first called and stored in a separate memory address even after the function has ended. (according to my understanding, after the function ended, all local variables should be freed?)
Am I correct?
In Python, functions are objects too, and the defaults are stored with the function object. Defaults are not locals; it is just that when the function is called, the arguments are bound to a default when not given an explicit value.
When Python encounters a def <functionname>(<arguments>): statement, it creates a function object for you there and then; this is 'definition time'; the function is not called but merely created. It is then that defaults are evaluated and stored, in an attribute on the function object.
Then when you call the function, the defaults have already been created and are used when you didn't provide a more concrete value for the argument. Because the defaults are stored with the function object, you get to see changes to mutable objects between function calls.
The locals are still cleared up of course, but as they are references (all identifiers in Python are), the objects they were bound to are only cleared up if nothing else is referencing them anymore either.
You can take a look a the defaults of any function object:
>>> def foo(bar='spam', eggs=[]):
... eggs.append(bar)
... return eggs
...
>>> foo.__defaults__
('spam', [])
>>> foo()
['spam']
>>> foo.__defaults__
('spam', ['spam'])
>>> foo() is foo.__defaults__[1]
True
The foo() function has a __defaults__ attribute, a tuple of default values to use when no values for the arguments have been passed in. You can see the mutable list change as the function is called, and because the function returns the eggs list, you can also see that it is the exact same object as the second value in that tuple.
If you don't want your defaults to be shared and instead need a new value for a parameter every time the function is called, but the parameter is not given, you need to set the default value to a sentinel object. If your parameter is still set to that sentinel in the function body, you can execute code to set a fresh default value. None is usually the best choice:
def foo(bar='spam', eggs=None):
if eggs is None:
eggs = []
If it should be possible to use None as a non-default value, use a singleton sentinel created beforehand:
_sentinel = object()
def foo(bar='spam', eggs=_sentinel):
if eggs is _sentinel:
eggs = []
The function that you have defined f is an object in its own regard. When you define defaults, these defaults are bound to the function that you have created.
You can see this in action:
>>> def f(a, L=[]):
... L.append(a)
... return L
>>> print id(f)
4419902952
>>> print f.__defaults__
([],)
>>> f(1)
[1]
>>> print id(f)
4419902952
>>> print f.__defaults__
([1],)
edit, further, you can see that the list container does not change either:
>>> print id(f.__defaults__[0])
4419887544
>>> f(2)
[1, 2]
>>> print id(f.__defaults__[0])
4419887544
On each subsequent call, the default list ("L") of your f function will have your a value appended.
A function is just an object in python, that is created using the def syntax. Default values are stored within the function object when the function is defined, and they are not re-evaluated later.
This is sometimes used to create function variables that persist to subsequent invocations. You can use the __defaults__ methods to check what the default values are for your function.
A common way to initialize new objects instead of reusing the same is:
def f(a, L=None):
if L is None:
L = []
L.append(a)
return L
You can check this page for more details.
Sorry this answer was meant for a different question, but I'll leave it here as a reference if anyone who wants to look at it. Define once means that at the first point when the code is executed, the default variable gets assigned to an object which is retained within the function object itself.
Notice only 1 object address gets printed, the default list object is used.
def f(a, L=[]):
print("id default: ", id(L))
L.append(a)
print("id used: ", id(L)
return L
Notice 2 different object addresses are printed, when you perform L=[] within the function, you are binding L to a different list object, therefore the default list object does not get change.
def f(a, L=[]):
print("id default: ", id(L))
if L == []:
L = []
L.append(a)
print("id used: ", id(L))
return L
The function above is basically the same as the one below except it uses the None object instead of a empty list object.
def f(a, L=None):
print("id default", id(L))
if L is None:
L = []
L.append(a)
print("id used: ", id(L))
return L
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Closed 10 years ago.
Possible Duplicate:
Creating lambda inside a loop
In the code below, invoking any member of the returned array of closures
prints the number 4.
def go():
x = []
for i in range(5):
def y(): print i
x.append(y)
return x
I would like each member of the closure to print the number that i was when the closure was defined.
One way around this is to use default arguments:
def y(i=i):
print i
Default arguments are evaluated when the function is created, not called, so this works as you'd expect.
>>> i = 1
>>> def y(i=i): print i
...
>>> i = 2
>>> y()
1
A little extra info just for fun:
If you're curious what the defaults are, you can always inspect that with the .func_defaults attribute (__defaults__ in python3.x):
>>> y.func_defaults
(1,)
This attribute is also writeable, so you can in fact change the defaults after the function is created by putting a new tuple in there.