At Thursday 2/11/2006 19:23, SpreadTooThin wrote:
I realize I may be beating a dead horse here... but...
def fn(x):
x = x + 1
print x
a = 2
fn(a)
>>> 3
print a
>>> 2
So in some cases the it is safe to assume that your variables to
function will not
change in other cases it is not.. but they are all the same...
Forget about variables!
In Python you have objects. Objects don't have a name, but you may
use a name to refer to them.
In the code above, `a` is not a variable, neither an object; it's
just a name referring to an object (a name bound to an object); such
object is an instance of type int, with value 2, to be precise.
fn(a) calls function fn [in fact, "the object bound to the name fn"]
passing an object -the object referenced by the name `a`- as its only argument.
Inside function fn, it gets a single argument bound to the name x.
You do some math and bind the resulting object to the local name `x`.
Now x refers to another instance of type int, with value 3. The
original int object is not modified - and never could have been,
since int's are immutable, you cant change them (a 2 will always be a
2: if you get a 3, it's ANOTHER object).
After function fn finishes, execution continues with the `print a`
statement. Nobody has changed the object pointed by the name `a`, so
it's still the int with value 2, and that's what you get.
All objects have identity. You can test identity using the `is`
operator: `a is b` returns True iff both objects are identical (both
are "the same object" just being referred by another name). The
function id() returns the object identity.
Some objects (by example, containers like lists and dicts) are
mutable. That means that you can change it's value, but the remain
being the same object. By example:
--- begin test.py ---
def g(x):
print 'x[1], in:', x[1]
x[1] = 10
print 'x[1], out:', x[1]
a1=1
a2=2
a3=3
foo=[a1,a2,a3]
print "before, foo=",foo,"id(foo)=",id(foo)
print "a2=",a2,"id(a2)=",id(a2)
g(foo)
print "after, foo=",foo,"id(foo)=",id(foo)
print "a2=",a2,"id(a2)=",id(a2)
--- end test.py ---
Output:
before, foo= [1, 2, 3] id(foo)= 12350728
a2= 2 id(a2)= 11163404
x[1], in: 2
x[1], out: 10
after, foo= [1, 10, 3] id(foo)= 12350728
a2= 2 id(a2)= 11163404
Function g modifies the second "slot" in the list - it does not
modify the object already in that place, it just makes the second
item in the list to refer to another object (the int 10). The
previous reference (the int 2) is lost.
Anyway, the list remains "the same" (look at id(foo)).
The name a2 still refers to the int 2 (because nobody changed that).
It's really easy once you get it - and then, just enjoy writing Python code!
--
Gabriel Genellina
Softlab SRL
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