We have at least found out that it has something todo with the timed loops (or timed structures). The stop button is read as F and the loop waits at the Dequeue node. Perhaps there is some issue with all these VIs being reentrant? i am sending the VI . http://softacoustik.com/labview-error/labview-error-6.php
I can see many examples of this happening in various parts of my code where queues either become invalid while waiting or are invalid when passed to a subVI, even though Showing results for Search instead for Did you mean: Reply Topic Options Subscribe to RSS Feed Mark Topic as New Mark Topic as Read Float this Topic to the Top Bookmark I have been using this code for a very long time (5+ years) without having a memory overrun issue.The problem I am having now is something is killing the queue reference Most errors on your machine are caused by uninstalling programs, installing new ones and accidentally deleting important files.
To stop the VI I need release queue, isn't it?4. BTW: This problem only happens in the EXE deployed to a target machine and only after running for several days. Once is enough and you can then use the error. The top VI (this VI) is a template that gets spawned many times.
Register a new account Sign in Already have an account? Members 0 1 post Posted October 16, 2008 Hi! Please try the request again. Show more post info Size: 1,178 bytes Customize: Reply 4: Re: why do i get error 1122 at dequeue element in consumer loop when i hit stop button Bob_Schor replied 1
Your cache administrator is webmaster. Or sign in with one of these services Sign in with Facebook Sign in with Twitter Sign in with LinkedIn Sign Up All Content All Content This Topic This Forum Advanced If memory is corrupted in shared clones, maybe named queues are better protected. /J Share this post Link to post Share on other sites PJM_labview 32 The 500 club Members https://forums.ni.com/t5/LabVIEW/Release-Queue-error/td-p/203122 Share this post Link to post Share on other sites John Lokanis 75 The 500 club Members 75 786 posts Location:Seattle, WA Version:LabVIEW 2015 Since:1993 Posted September 19, 2008 I
Sign In Sign In Remember me Not recommended on shared computers Sign in anonymously Sign In Forgot your password? nathand Trusted Enthusiast 05-27-2013 10:06 PM Options Mark as New Bookmark Subscribe Subscribe to RSS Feed Highlight Print Email to a Friend Report to a Moderator There's nothing wrong, you just How do I troubleshoot it?So here's my program, read-table.viAnd there's another thing about multicolumn listbox put inside main vi.But I need to create a subvi (my current read_table.vi) that receives the Producer consumer loop : Write waveform data to consumer loop Release notifier outside of consumer loop Release notifier outside of consumer loop Loop Termination/Camera Reinitialization in Master/Slave...
I do not 'obtain' an existing quene anywhere because I am using unnamed queues. You should not need to destroy the queues, just close all the references you open. (there are good times to set destroy=True, but don't just set it because you feel like Are you sure your second while loop can process the data as fast as it is acquired? ToeCutter 1 user's latest post: why do i get error 1122 at...
I would like to program such that the sub-vi will process data and input to main.vi thru multicolumn listbox in REAL TIME, NOT after the subvi has finished before putting all http://softacoustik.com/labview-error/labview-error-ni-488.php I could zip up the code and send it to them, but they would not be able to run it. Your consumer loop should actually be releasing the queue in the STOP case. 3. Solution: This error commonly appears when one of the previously listed Queue functions or operations tries to operate on a reference which has been closed.
thanks, -John Hi John, I'm not sure if the following may be what is hitting you but you did ask for "other ideas" When a VI is no longer running, all All sorted now. Wire the release queue function out of the loop where you generate the events instead of the consumer loop. check over here Since you pressed the stop button, it is now T and the loop stops.
When the spawned VI finishes execution, it will leave memory, as will all of its queues, notifiers, etc.... Regarding the launcher, no, that part of the app continues to run even after the error. Your wait in the producer loop shold only be when there is no data in the buffer...
The easiest way is probably to use the the General Error Handler function in the Dialog & Error palette. This error is inevitable? XML Loop inside Loop(Set this element as loop, child node and parent will lose loop) New consumer desktops, consumer all-in-ones, and c... If it was memory corruption, then what could be causing it?
not only would you have to allocate millions of queues, you'd have to have them all continuously in play in order for the refnums to ever hit up against each other. Basic dataflow. Since the queue is unnamed, it is supported to be technically impossible for the refernce to go invalid while the VI is running and the release queue has not been called. this content Started by John Lokanis, August 30, 2008 41 posts in this topic Prev 1 2 Next Page 1 of 2 John Lokanis 75 The 500 club Members 75 786
We get error 1122 all the time, because we kill the queues on purpose to stop our processes, but it never happens unexpectedly. :thumbdown: Violence is a bad thing. In my case, however, I don't think that is possible. Show more post info Size: 235 bytes Customize: Reply 5: Re: why do i get error 1122 at dequeue element in consumer loop when i hit stop button matt198717 replied 1 Sign in here.
At step 2 we wait for messages to appear. In my case, however, I don't think that is possible. What it boils down to is LabVIEW is corrupting its own memory when running large parallel apps with a lot of shared clone Vis. ie are you running out of CPU cycles?
Share this post Link to post Share on other sites mprim 0 Active Members 0 19 posts Version:LabVIEW 2009 Since:2009 Posted September 24, 2008 Your vi makes a number of