OK David,
first of all thanks for your precious help...
only a clarifying question (clarifying for me of course... :)):
when the library function execution exceeds the timeout value,
the piece of code in:
if(!semimutex.wait(timeout)) {
----------> library must have crashed, etc...
}
runs automatically? That is: is there a sort of "goto" that let the
execution continue within the
if statement? I put this question because (besides my ignorance), if I
have to put an external loop that checks for the
timeout, like...
if(isPending(pendingInput)) {
---> while(1) {
if(!semimutex.wait(timeout)) {
library must have crashed, etc...
}
call library function ...
semimutex.post();
}
}
it probably won't resolve my problem, because when external library get
stucked, the program can't go on,
and never will reach the beginning of the external loop, never will
excute the piece of code of the timeout condition.
If there is an automatic "goto" instead, then I think it will work fine.
And more, in this way I can implement a sort of "time sharing", because
even if my server is multithread, actually only 1 thread can execute
until
the library function exits (it isn't a fix execution time), so it's a
fake multithread and performance speed is heavily penalized indeed.
To have this working, I have to increase the couter of the semaphore?
Can you make other example of the semaphore usage please?
or suggest some documentation to read about this?
Thanks again for your help.
Carlo
David Sugar ha scritto:
What you are thinking of is something like this:
Semaphore semimutex(1);
...
if(isPending(pendingInput)) {
if(!semimutex.wait(timeout)) {
library must have crashed, etc...
}
call library function ...
semimutex.post();
}
Hence wait is used like a enterMutex, post like a leaveMutex, and only
one instance can pass through because it has a count of 1. Since wait
supports a timeout, you can use that for your failure detect.
The Common C++ Semaphore is actually implemented as a combination of
mutex and a conditional.
Wakan wrote:
Thanks for your reply.
Can you write an example with the semaphore in this context?
Thanks,
Carlo
Wakan ha scritto:
Thanks for your reply.
Can you write an example with the semaphore in this context?
Thanks,
Carlo
David Sugar ha scritto:
I think what he really wants to do is use the Common C++ Semaphore with
a count of 1, since semaphore can block with a "timeout", and with a "1"
count in effect acts as a mutex...
The larger question though of what to do with the stuck thread, perhaps
from a broken library, is important also. He could cancel the thread in
question if it timed out, and put cleanup code in the thread destructor
when he invokes delete to join it, but if the library allocated
resources he cannot determine for them to get cleanly deallocated.
Conrad T. Pino wrote:
You're on the right track. Please pardon my rusty recollection syntax.
Revise:
mutex.enterMutex();
to a form that accepts a timeout. This example won't compile:
const int timeout = 500;
while (1) {
if ( isPending( pendingInput ) ) {
int result = mutex.enterMutex( timeout );
if ( result == success ) {
---> protected function call (a function that can't work
in multithread env) <---
mutex.leaveMutex();
} else {
// do something else with pending input
}
}
}
While the above will implement your timeout idea it does nothing about
the fundamental problem. When the library "breaks" can be discoverd
and the blocked mutex/thread combination can be cancelled/restared
HOWEVER PROPRIETARY LIBRARY STATE IS INCONSISTENT & LIKELY DANGEROUS.
If the library is dynamically loaded under your control then unload it
and reload it to get it back into a consistent state. If it's auto or
static linked then you may have an unsolveable problem.
If a reliable system is your goal then one of these MUST be true:
proprietary library NEVER fails
OR
proprietary library can be forced back to a consistent
state because it has a reload or reinitialize method
Good luck.
-----Original Message-----
From: address@hidden
[mailto:address@hidden]On Behalf Of Wakan
Sent: Saturday, January 27, 2007 07:18
To: address@hidden
Subject: can't release mutex
Hi,
I'm using commoncpp library in a project that includes a multithread
server that
can accept many tcp connection and each connection generates a thread.
In this server I'm using a thirdy party proprietary library (compiled),
that (as the producer says)
can't work in multithread environment.
As each connection thread needs to call functions of this thirdy party
library, I've used a mutex to protect the piece of
code where this external library function is called.
In this way I can use this library fine, because all calls are
serialized and the protected function is called by one thread each time.
This is an example:
...
while(1) {
if(isPending(pendingInput)) {
mutex.enterMutex();
---> protected function call (a function that can't work
in multithread env) <---
mutex.leaveMutex();
}
}
...
in 99% of cases it works fine. But it accidentally happens that a thread
can't exit from the mutex
maybe because the thirdy party function freezes itself.
When this happens, the server can accept other connections, soit still
works, but each thread can't enter in the mutex proteced piece of code.
I'm thinking about something, like a timeout, that exits the mutex after
a time, if the thread is not exiting...
Can someone help me to resolve this problem? Are there other solutions?
Thanks in advance,
Regards
Carlo
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