Kent Johnson wrote:
On Mon, Mar 16, 2009 at 12:30 PM, A.T.Hofkamp <a.t.hofk...@tue.nl> wrote:
I don't know what code is executed in an assignment exactly, but
**possibly**, first the 'read()' is executed (thus loading a very big string
into memory), before assigning the value to the variable (which releases the
previous value of the variable).
That means that just after reading but before assigning, you **may** have
two very big strings in memory that cannot be garbage collected.
No. Python variables are references to values, not containers for
values. Python assignment is *always* reference assignment, not
copying of a value. More here;
http://personalpages.tds.net/~kent37/kk/00012.html
Nice web-page!
I am aware of how variables are treated in Python.
Let me explain my reasoning in more detail. Consider the statement
s = s + "def"
under the assumption that s is now "abc" (or rather, s references the data
value "abc").
For the assignment, I believe the following happens inside the python
interpreter (but I am not 100% sure that it is exactly the sequence):
1. get a reference to the current value of s.
2. get a reference to the constant value "def".
3. compute the new value "abcdef", store it in memory, and make a reference to
it.
4. drop the old reference of s (thus free-ing "abc")
5. give s a reference to the newly computed value.
The point I was trying to make is that after step 3 and before step 4, the old
value of s is still referenced by s, and the new value is referenced
internally (so step 5 can be performed).
In other words, both the old and the new value are in memory at the same time
after step 3 and before step 4, and both are referenced (that is, they cannot
be garbage-collected).
Assuming that the above mechanism is used with all assignments, this will also
be the case in the sequence of assignments
s = read()
# write s
s = read()
# write s
where in the second assignment statement, the read() is done (creating the new
value in memory) before dropping the value of s from the first assignment.
You can do the above statements also iteratively of course
for i in ...
s = read()
# write s
but since the loop does nothing with either s or read(), this will not change
how the assignment works.
In the case that you are manipulating large values (as in taking a lot of
computer memory for each value), the execution of the read() during step 3 may
fail due to memory being used for the previously assigned value of s.
Sincerely,
Albert
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