On 2023-01-13 16:57:45 +0100, Jen Kris via Python-list wrote: > Thanks for your comments. You make a good point. > > Going back to my original question, and using your slice() example: > > middle_by_two = slice(5, 10, 2) > nums = [n for n in range(12)] > q = nums[middle_by_two] > x = id(q) > b = q > y = id(b) > > If I assign "b" to "q",
You don't asssign b to q. You assign q to b. Assignment is not
commutative, the direction matters.
> then x and y match – they point to the same memory until "b" OR "q"
> are reassigned to something else.
Correct.
> If "q" changes during the lifetime of "b" then it’s not safe to use
>the pointer to "q" for "b", as in:
There is no pointer to q[1]. q is a pointer to something (an object of type
list with 3 elements).
b is a pointer to the same object at this point. Of course, if you
assign a different pointer to q, then q will point to the new object
from that point on while b will continue to point to the original object
(until it is also re-assigned).
> nums = [n for n in range(2, 14)]
> q = nums[middle_by_two]
> x = id(q)
> y = id(b)
>
> Now "x" and "y" are different, as we would expect.
> So when writing a spot speed up in a compiled language,
If you are writing a Python compiler you certainly will have to adjust
for the fact that the same variable may hold pointers to very different
objects over its lifetime. But as far as I understand it, you aren't
trying to write a Python compiler, just an extension, so that shouldn't
concern you.
> The motivation behind my question was that in a compiled extension
> it’s faster to borrow a pointer than to move an entire array if it’s
> possible, but special care must be taken.
I still don't understand what you mean by that.
hp
[1] Not quite true. The run-time system must keep track of the
variables, so there likely is a pointer to q somewhere. But it's (IMHO)
irrelevant to this discussion, unless the extension is trying to look up
variables by name or something like that.
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| | | h...@hjp.at | -- Charles Stross, "Creative writing
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