Am 02.11.2010 13:51, schrieb John:
Hello, I thought the following should end with G[1] and G[0] returning
20. Why doesn't it?
In [69]: a = 10
"a" is a reference to an immutable object of type int with value 10.
Somewhere in memory is an object of type int with value 10 stored. This
object cannot be changed. "a" is just a name that references to the
memory adress.
In [70]: G = {}
In [71]: G[0] = [a]
now, G[0] holds a reference to a mutable object of type list with an
immutable object of type int with value 10 as its content. So, your list
holds as item the memory adress of an object int with value 10. Your
list knows nothing about other names (like "a"), that refer to the same
adress in memory. You have handed over the object to the list (or memory
adress of the object), not the name "a".
In [72]: G[1] = G[0]
In [73]: a = 20
Now, you've created a *new* immutable object of type int with value 20.
"a" now refers to the adress of the new integer object. "a" has
forgotten everything about its former reference to the object of type
int with value 10.
You have *not* changed the "old" object of type int with value 10. Your
list in G[0] still holds the memory adress of the first int with value 10.
In [74]: G[1]
Out[74]: [10]
In [75]: G[0]
Out[75]: [10]
??
Now, if you would keep a reference to the list you've handed over to the
dict, you could change the content of the list, as the list is mutable.
>>> a = [10]
>>> G = {}
>>> G[0] = a
>>> G[0]
[10]
>>> a[0] = 20
>>> G[0]
[20]
Take care of some operations that return a copy of the list:
>>> a = a[:]
Now, "a" refernces to a new list, which has also one item of type in
with value 20. But it's not the same list as originally!
>>> a[0] = 30
Changing this new list ...
>>> G[0]
[20]
does not affect the list, we stored in the dict.
HTH,
Jan
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