bizarre id() results

Stuart McGraw smcg4191zz at friizz.RimoovAllZZs.com
Thu Dec 15 22:19:15 CET 2005


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.





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