bizarre id() results

[Stuart McGraw]

The following was cut and pasted exactly (except for the
# lines which I added after the fact) from an interactive python
session in a Window 2000 cmd.exe window.

Can somebody please explain to me what the heck is
going on?!?!

Python 2.4.2 (#67, Sep 28 2005, 12:41:11) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright" , "credits" or "license" for more information.

class A: .... def m1(self): print "m1"
.... def m2(self): print "m2"
.... a = A()
a.m1() m1 a.m2() m2
# ok, both methods work and give the expected results
# so i presume they are different methods. id(a.m1) 9202984 id(a.m2) 9202984 id(a.m1)==id(a. m2) True
# Huh? They seem to be the same. a.m1 is a.m2 False
# But not the same... a.m1 <bound method A.m1 of <__main__.A instance at 0x00923B98>> a.m2 <bound method A.m2 of <__main__.A instance at 0x00923B98>>
# Let's look at them in hex... hex(id(a.m1)) '0x8c6d28' hex(id(a.m2)) '0x8e7b48'
# Now they are different. 0x8c6d28->9202984, 0x8e7b48->9337672 id(a.m1) 9337672 id(a.m2) 9337672
# Now they are both equal to the second one. hex(id(a.m1)) '0x8e7b48' hex(id(a.m2)) '0x8e7b48'
# in hex too. id <built-in function id> hex

<built-in function hex>
# just double checking!

Why??? This is so bizarre I'm sure I am doing something
really stupid.

[Diez B. Roggisch]

> # ok, both methods work and give the expected results

# so i presume they are different methods.

id(a.m1)
9202984
id(a.m2)
9202984
id(a.m1)==i d(a.m2)

True
# Huh? They seem to be the same.

What you observe is rooted in two things:

- python objects bound methods are created on demand. Consider this
example:
class Foo:
def m(self):
pass

f = Foo()
m1 = f.m
m2 = f.m

print id(m1), id(m2)
The reason is that the method iteself is bound to the instance - this
creates a bound method each time.

The other thing is that this bound method instances are immediaty
garbage collected when you don't keep a reference. That results in the
same address being used for subsequent bound method instantiations:

print id(f.m)
print id(f.m)
results in the same id being printed. It doesn't matter if you use m1,
m2 instead, as in your example.

HTH,

Diez

[Fredrik Lundh]

Stuart McGraw wrote:

The following was cut and pasted exactly (except for the
# lines which I added after the fact) from an interactive python
session in a Window 2000 cmd.exe window.

Can somebody please explain to me what the heck is
going on?!?!

Python 2.4.2 (#67, Sep 28 2005, 12:41:11) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright" , "credits" or "license" for more information.

class A: ... def m1(self): print "m1"
... def m2(self): print "m2"
... a = A()
a.m1() m1 a.m2() m2
# ok, both methods work and give the expected results
# so i presume they are different methods. id(a.m1) 9202984 id(a.m2) 9202984 id(a.m1)==id(a. m2) True
# Huh? They seem to be the same. a.m1 is a.m2 False
# But not the same... a.m1 <bound method A.m1 of <__main__.A instance at 0x00923B98>> a.m2 <bound method A.m2 of <__main__.A instance at 0x00923B98>>
# Let's look at them in hex... hex(id(a.m1)) '0x8c6d28' hex(id(a.m2)) '0x8e7b48'
# Now they are different. 0x8c6d28->9202984, 0x8e7b48->9337672 id(a.m1) 9337672 id(a.m2) 9337672
# Now they are both equal to the second one. hex(id(a.m1)) '0x8e7b48' hex(id(a.m2)) '0x8e7b48'
# in hex too. id <built-in function id> hex

<built-in function hex>
# just double checking!

Why??? This is so bizarre I'm sure I am doing something
really stupid.

try running this script:

class bound_simulator :
def __init__(self, method):
self.method = method
print "alloc", method, id(self)
def __del__(self):
print "release", self.method, id(self)

class instance_simula tor:
def __getattr__(sel f, method):
return bound_simulator (method)

i = instance_simula tor()

print id(i.m1)
print id(i.m2)
print id(i.m1) == id(i.m2)
print i.m1 is i.m2

and see if you can figure out what's going on here.

</F>

[Chris Lambacher]

a.m1 returns a bound method which gets freed before you try checking a.m2,
which ends up getting the same peice of memory. If you save a reference to
the bound methods, they are forced to have separate objects.

class A: .... def m1(self): print "m1"
.... def m2(self): print "m2"
.... a = A()
a.m1() m1 a.m2() m2 id(a.m1) 25402184 id(a.m2) 25402184 b = a.m1
c = a.m2
id(b) 25402184 id(c)

25541848
-Chris

On Thu, Dec 15, 2005 at 02:19:15PM -0700, Stuart McGraw wrote:

The following was cut and pasted exactly (except for the
# lines which I added after the fact) from an interactive python
session in a Window 2000 cmd.exe window.

Can somebody please explain to me what the heck is
going on?!?!

Python 2.4.2 (#67, Sep 28 2005, 12:41:11) [MSC v.1310 32 bit (Intel)] on win32
Type "help", "copyright" , "credits" or "license" for more information. class A: ... def m1(self): print "m1"
... def m2(self): print "m2"
... a = A()
a.m1() m1 a.m2() m2
# ok, both methods work and give the expected results
# so i presume they are different methods. id(a.m1) 9202984 id(a.m2) 9202984 id(a.m1)==id(a. m2) True
# Huh? They seem to be the same. a.m1 is a.m2 False
# But not the same... a.m1 <bound method A.m1 of <__main__.A instance at 0x00923B98>> a.m2 <bound method A.m2 of <__main__.A instance at 0x00923B98>>
# Let's look at them in hex... hex(id(a.m1)) '0x8c6d28' hex(id(a.m2)) '0x8e7b48'
# Now they are different. 0x8c6d28->9202984, 0x8e7b48->9337672 id(a.m1) 9337672 id(a.m2) 9337672
# Now they are both equal to the second one. hex(id(a.m1)) '0x8e7b48' hex(id(a.m2)) '0x8e7b48'
# in hex too. id <built-in function id> hex

<built-in function hex>
# just double checking!

Why??? This is so bizarre I'm sure I am doing something
really stupid.
--
http://mail.python.org/mailman/listinfo/python-list