Problem with Functions which has default parameter(s)

I have some question about functions which have default parameters.

import sys
from random import randint

my_list = [1,2,3]
def Veli(a, b = my_list):
    my_list.append(randint(1,1500))
    print a + b[-1]

print "Number of Refs:", sys.getrefcount(my_list)
print "Def args:", Veli.func_defaults
Veli(1) # This is 4
Veli(1) # We almost always obtain different result because of randint
Veli(1) # Again different results.
print Veli.func_defaults
print "Number of Refs:", sys.getrefcount(my_list)
Veli(1)
my_list = [-5] # old my_list has different address 
print "Number of Refs:", sys.getrefcount(my_list) # Less than previous
print "Def args:", Veli.func_defaults
Veli(1) # Now gives same results.
print "Def args:", Veli.func_defaults
Veli(1) # Same result again...

Outputs: (Some numbers depend on which values randint has returned, of course.)

Number of Refs: 3

Def args: ([1, 2, 3],)

322

1119

740

([1, 2, 3, 321, 1118, 739],)

Number of Refs: 3

303

Number of Refs: 2

Def args: ([1, 2, 3, 321, 1118, 739, 302],)

303

Def args: ([1, 2, 3, 321, 1118, 739, 302],)

303

The following code gives you a number less than the previous one, because b and my_list is not referenced to same address anymore.

print "#ref:", sys.getrefcount(my_list) # Less than previous

Now, we have a way (the only way?) to reach b, default argument of the function:

Veli.func_defaults[0]

Sorry about my long explanation. Here is my questions:

1. Is this a problem? Dictionary of my module crush my_list variable then now my_list variable has default address than previous. Then function which uses global variable named my_list in its body is changing (growing) while my default argument is not. Why cannot b see global variable named my_list when a + b[-1] expression is executed? I know, b has different address with my_list (because mutual objects (like list) are guaranteed to refer to different, unique, newly created list) now, but why were Python implemented so that b cannot see the global variables when b is a function arguments? Could you explain comprehensively?

2. Is there any way to get same results with Veli.func_defaults[0]? 开发者_如何学CThis code executed then let say, I want to change my default arguments of my function named Veli. I cannot do this with my_list, because my_list and b have different addresses. One way is changing the elements of the list, Veli.func_defaults[0]. Is there any different way(s)?

3. (It is not so related to code above) How can get addresses of variable? For example, how can get address of b? I use built-in function such as __hash__ , but there should be more appropriate way.

Notes:

a) This codes may be useless for any reason, but I want to learn opinions.

b) Python 2.6.6 (r266:84292, Sep 15 2010, 15:52:39)

[GCC 4.4.5] on linux2

1. Defaults arguments are bounds at function definition time. The line def Veli(a, b = my_list): puts a reference to the object my_list happens to refer to at that time into func_defaults. Python never uses pass-by-reference - variables hold references and those references are always passed by value. So what's actually saved in b, a reference (pointer), is copied and nobody remembers where it came from or bothers to update it.
2. Not quite sure what you're asking, please clarify. b will be Veli.func_defaults[0] if no second argument was passed, but obviously different if it is passed (well, unless of course the caller accesses func_defaults... you should assume he doesn't).You could, as suggested in another answer, make bdefault to None and use the global my_list if b is None - this would give you a fresh, updated copy of its reference on every call. Perhaps you should write a class and keep the default value as an attribute of self and apply the usual (... = None): if ... is None: use = default idiom.
3. The build-in function id. Actually, it's implementation defined what this returns (doesn't have to be the address), as long as it's an integer that represents the object identity, i.e. distinct objects (with overlapping lifetimes) have a distinct id and the same object always gives the same id during its lifetime. The easy return value, which CPython chooses, is the address.

I suggest you look at this existing question, as your question is essentially related to the same problem (albeit with a slightly different perspective on the topic).

As you note, when the function definition is executed, b is bound to point to my_list. When you later change my_list to point to a different object, b remains unaffected.

To get "lazy binding" you want to write your function like this:

def Veli(a, b=None):
    if b is None:
        b = my_list # Gets value of my_list at call time
    my_list.append(randint(1,1500))
    print a + b[-1]