On Oct 28, 3:29 pm, jasiu85 <[EMAIL PROTECTED]> wrote:
> On Oct 27, 10:12 am, "Diez B. Roggisch" <[EMAIL PROTECTED]> wrote:
>
>
>
> > jasiu85 schrieb:
>
> > > Hey,
>
> > > Please take a look at the code of the two threads below:
>
> > > COMMON_DICT = {}
>
> > > def thread_1():
> > > global COMM
On Oct 27, 10:12 am, "Diez B. Roggisch" <[EMAIL PROTECTED]> wrote:
> jasiu85 schrieb:
>
>
>
> > Hey,
>
> > Please take a look at the code of the two threads below:
>
> > COMMON_DICT = {}
>
> > def thread_1():
> > global COMMON_DICT
> > local_dict = prepare_dict()
> > COMMON_DICT = local
jasiu85 wrote:
> Do I need a lock to protect the COMMON_DICT dictionary? AFAIK bytecode
> operations are atomic and in each thread there's only one crucial
> bytecode op: STORE_NAME in the first thread and LOAD_NAME in the
> second one. So I suspect that everything will work just fine. Am I
> righ
jasiu85 <[EMAIL PROTECTED]> wrote:
> Hey,
>
> Please take a look at the code of the two threads below:
>
> COMMON_DICT = {}
>
> def thread_1():
> global COMMON_DICT
> local_dict = prepare_dict()
> COMMON_DICT = local_dict
>
> def thread_2():
> global COMMON_DICT
> local_dic
jasiu85 schrieb:
Hey,
Please take a look at the code of the two threads below:
COMMON_DICT = {}
def thread_1():
global COMMON_DICT
local_dict = prepare_dict()
COMMON_DICT = local_dict
def thread_2():
global COMMON_DICT
local_dict = COMMON_DICT
use_dict(local_dict)
Do
Hey,
Please take a look at the code of the two threads below:
COMMON_DICT = {}
def thread_1():
global COMMON_DICT
local_dict = prepare_dict()
COMMON_DICT = local_dict
def thread_2():
global COMMON_DICT
local_dict = COMMON_DICT
use_dict(local_dict)
Do I need a lock to pr
