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Labview Error 1122

I am able to interact with it even after the error is reported. In order to get any recycling of the unnamed queue IDs you would not only have to generate roughly 30 million queues, you would also need to get particularly (un)lucky on The code relies on a continuous flow of data between an in-house SQL server and the application. That's your two 0's. ___________________Try to take over the world! 2 Kudos Message 8 of 11 (2,836 Views) Reply 2 Kudos Re: Release Queue error vicens Member ‎07-06-2007 03:01 AM Options this content

What appears to be happening is when some reentrant or VIT spawned code leaves memeory, it accidentally kills other queues in other instances of the same VIs. Show more post info Size: 275 bytes Customize: Reply 8: Re: why do i get error 1122 at dequeue element in consumer loop when i hit stop button crossrulz replied 1 I do not understand. Since it is impossible for these error to occur due to a coding issue, there must be a bug in LabVIEW causing it to destroy a queue reference outside this process. http://digital.ni.com/public.nsf/allkb/10AE5BC4A92626058625780600749827

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. Since all of these VIs are part of the main VI, i don't see how it is possible that the queue reference would be automatically removed from memory. So, there is no way that cleanup VI could execute before the VI that is waiting. 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

All Rights Reserved.Designated brands and trademarks are property of their respective owners.Use of this website constitutes acceptance of BoardReader's Terms of Service and Privacy Policy. 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 The work-around it to make sure the VI's that creaed the queue don't go idle until after the queues is destroyed. Share this post Link to post Share on other sites PJM_labview 32 The 500 club Members 32 777 posts Version:LabVIEW 2009 Since:1998 Posted September 19, 2008 Just to make sure,

The Wait generates Error 1122 and stops the loop and closes the VI. Is there a possibility the launcher is going idle? But, I have searched all the VIs and the only one where the queue is destroyed is in the cleanup VI that comes after this VI and is connected by the https://forums.ni.com/t5/LabVIEW/Release-Queue-error/td-p/203122 PJM Share this post Link to post Share on other sites John Lokanis 76 The 500 club Members 76 786 posts Location:Seattle, WA Version:LabVIEW 2015 Since:1993 Posted September 20, 2008

I have a large number of reentrant VIs running and I create a lot of unnamed queues that I pass inside a cluster to sub VIs. Show more post info Size: 345 bytes Customize: Reply 7: Re: why do i get error 1122 at dequeue element in consumer loop when i hit stop button matt198717 replied 1 The structure of my code has a main vi that calls a sub VI to create the queue and then passes the queue ref to another sub VI that listens to The queue was destoryed, so you get error 1 and 0.

But, I have searched all the VIs and the only one where the queue is destroyed is in the cleanup VI that comes after this VI and is connected by the useful source There's only one Release Notifier in the application that can only run when the operator Exits the application. That seems unlikely. http://lavag.org/old_files/monthly_08_2008/post-2411-1220050049.jpg' target="_blank"> The Dequeue element gets an error stating the reference has become invalid while waiting.

Answered Your Question? 1 2 3 4 5 Document needs work? news But if you're convinced there's a bug in LV and only a huge application replicates it, then, please, submit the whole app if you can. For example if the input is a path, the path might contain a character not allowed by the OS such as ? I only use the shared clones mode, but none of them have a uninitialized shift register...

Remove the "wait for ms multiple" function and wire a constant to the timeout of the dequeue function 2. Regarding the launcher, no, that part of the app continues to run even after the error. I am trying to do a loop in C. http://ascadys.net/labview-error/labview-error-10.html Other options include using global variables and functional global variables. ___________________Try to take over the world! 1 Kudo Message 2 of 11 (3,170 Views) Reply 1 Kudo Re: Release Queue error

This report page is a snippet summary view from a single thread "why do i get error 1122 at dequeue element in consumer loop when i hit stop button", located on Ben Share this post Link to post Share on other sites John Lokanis 76 The 500 club Members 76 786 posts Location:Seattle, WA Version:LabVIEW 2015 Since:1993 Posted August 30, 2008 I want to ensure that no data is missed so the top loop has to run as fast as possible to avoid a buffer overrun.

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Edit: When I change the program as in the attachment, the error message does not pop-up. Sign In Sign In Remember me Not recommended on shared computers Sign in anonymously Sign In Forgot your password? Show more post info Size: 100 bytes Customize: Reply 3: Re: why do i get error 1122 at dequeue element in consumer loop when i hit stop button crossrulz replied 1 The LV-functions that causes the application to freeze are all inside a timed loop.

Mark QUOTE (jlokanis @ Sep 23 2008, 10:57 AM) Thanks for the observations. Sign In Now Sign in to follow this Followers 0 Go To Topic Listing LabVIEW Bugs All Activity Home Community LabVIEW Feedback for NI LabVIEW Bugs What can kill a queue? 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 check my blog I have a sneaking suspicion that there are some latent bugs in the queue feature.

It would be great if I could find something I screwed up that is causing this, but I can't think of a single thing that could. I notice that when this problem occurs, the whole app also starts to slow down AND memory usage starts to increase. When the listener quits, it passes its error cluster to the sub VI that destroys the queue. I do not understand.

Are you sure your second while loop can process the data as fast as it is acquired? Search Advanced search Search everywhere only in this thread Thread: why do i get error 1122 at dequeue element in consumer loop when i hit stop button Started 1 Your consumer loop should actually be releasing the queue in the STOP case. 3. newbieeng Active Participant ‎05-28-2013 12:29 PM Options Mark as New Bookmark Subscribe Subscribe to RSS Feed Highlight Print Email to a Friend Report to a Moderator Ok thanks.

So, there is no way that cleanup VI could execute before the VI that is waiting. That seems unlikely. Showing results for  Search instead for  Did you mean:  Reply Topic Options Start Document Subscribe to RSS Feed Mark Topic as New Mark Topic as Read Float this Topic to the Maybe it is time to move up to 8.6?

Is there a way? 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 Sign in to follow this Followers 0 What can kill a queue? Here is an image of my code: Looking at the image, at step 1, an unnamed queue is created.

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