Pierre Barbier de Reuille wrote:
>Great :)
>
>Just to be clear about that: you can see the "connect" line as a dynamic
>registration process (with the symetric disconnect operation available
>via ... "disconnect"). The only thing you need is to connect (at
>runtime) the even before it's called (ob
Great :)
Just to be clear about that: you can see the "connect" line as a dynamic
registration process (with the symetric disconnect operation available
via ... "disconnect"). The only thing you need is to connect (at
runtime) the even before it's called (obvious isn't it ? ;) ), but you
have no o
Pierre Barbier de Reuille wrote:
>nephish a écrit :
>
>
>>one more thing.
>>if i uncomment the lines
>>gtk.threads_enter()
>>and
>>gtk.threads_leave()
>>the whole thing locks up when the function is called.
>>the gui, and the thread both lock up.
>>
>>
>
>Well, that's just normal. However, w
Pierre Barbier de Reuille wrote:
>nephish a écrit :
>
>
>>one more thing.
>>if i uncomment the lines
>>gtk.threads_enter()
>>and
>>gtk.threads_leave()
>>the whole thing locks up when the function is called.
>>the gui, and the thread both lock up.
>>
>>
>
>Well, that's just normal. However, w
nephish a écrit :
>
> one more thing.
> if i uncomment the lines
> gtk.threads_enter()
> and
> gtk.threads_leave()
> the whole thing locks up when the function is called.
> the gui, and the thread both lock up.
Well, that's just normal. However, what you should do is to send a
signal from your th
nephish wrote:
> Serial1() just call the function ... it will be evaluated and then the
> result will be sent to Thread.start ...
>
> So try:
>
> Thread.start(Serial)
>
>
>ok,
>so far the thread runs fine in the background, but just does not send the
>output to the
>textb
Serial1() just call the function ... it will be evaluated and then the
result will be sent to Thread.start ...
So try:
Thread.start(Serial)
ok,
so far the thread runs fine in the background, but just does not send the
output to the
textbuffer, and hangs up th
