On Fri, May 28, 2021 at 4:14 AM Shreyan Avigyan
<pythonshreya...@gmail.com> wrote:
>
> > A context switch can happen between any two of those instructions.
> > That means one thread could load the global, then another thread could
> > load the same value, resulting in both of them writing back the same
> > incremented value. Or, between opcodes 6 and 8 (between the lines of
> > Python code), you could store the value, then fetch back a different
> > value.
>
> I see now. Then we can go with Steven's idea. Let's keep the changes in
> locals temporarily and when it yields or returns then modify the __statics__
> member. And even if it's a generator it will stop iteration some time and if
> it doesn't then the member wasn't meant to be modified.
>
So you're saying that...
def f():
static x = 0
x += 1
yield x
next(f())
next(f())
next(f())
will yield 1 every time? According to the "write-back later"
semantics, this is only going to actually update the static once it
gets nexted the second time. This would be utterly unique in all of
Python: that assignment doesn't happen when you say it should happen,
it instead happens arbitrarily later.
I would have to call that behaviour "buggy". If you use literally any
other form of persistent state, ANY other, it would increment.
Even worse: I don't see how your conclusion relates to my explanation
of threading, which you describe. Instead of having a minor risk of a
context switch in the middle of a statement (which can be controlled
with a lock), you have a major risk of a context switch ANYWHERE in
the function - and now there's no way you can use a lock to protect
it, because the lock would (by definition) be released before the
function epilogue.
ChrisA
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