Question about generating a delay :-P

Hello.
I have some movie clips on my root, and in a button press they all do something.
I want them to do that thing not immediately, but only after a delay.
What is the simplest way to delay the commands? setDelay or setTimer thing?
10ks.

Try using setTimeout()... it's explained in the help docs.

Similar Messages

Simple question about mount and delay script

Hello, I have a simple question about mounting volumes on start up.
My computer wakes up auto - and then auto there are serveral applescript tasks (mount and start up programs)
I use this script (daily) for serveral connections with external volumes I always need the connect to. ( In the script I use this code 3 times for other locations) I do this on start up.
set Uname to "XXX"
set Pword to "XXX"
set someVolume to "afp://XXX.XXX.XXX.XXX/XXX/XXX
mount volume someVolume as user name Uname with password Pword
(same code for 2 other locations)
When i do this on start up sometimes the script says there is no connection possible. I guess it's because on start up the connection isn't there. And 1 minute later when computer is total ready the connection is ok. When I run the script then It works. Just sometimes ( In the morning) 'It' want to connect but the script stops. When i do it manually(I run the script again) it works just fine.
Is it possible that I need a delay? Can someone explain this?
How would I make a delay handler for this script? Is that the best solution?
Thanks in advance. This is something small i'm wondering about.
Colin

BTW, If you saved the script as an application +(and not as an application bundle)+ you could drop the script on to *Drop Script Backgrounder* (freeware).
Then the script would run in the background, so, you wouldn't see it running in the Finder.
<http://www.macupdate.com/info.php/id/7922/drop-script-backgrounder-x>
Tom

Generic question about generating & saving XML

I need to write a process that will generate an invoice as XML, then store this entire XML document in a CLOB. I will have multiple stored functions and/or procedures that build various sections of the invoice. The process needs to be flexible enough to allow new functions to be added and old ones removed at any time with little impact.
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one solution i can think of is to use SQLX operator instead of dbms_xmlgen.
here is a sample example.
declare
l_xmltype xmltype;
l_deptno emp.deptno%type;
begin
for i in (select * from emp order by deptno)
loop
 select xmlconcat(
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 ,xmlelement("detpno", i.deptno))
 into l_xmltype from dual;
 dbms_output.put_line(l_xmltype.GetClobVal());
end loop;
end;
/Now here you can open the query once, keep writing to the file till the deptno
is same, when the deptno changes, close the file and open a new file with new
deptno and start writing.
Note : in this way you will have to add the xmlprolog manually to each of the file which should not be an issue. after opening the file add the prolog string manually.
Hope this helps.

Simple question about generate proxy

Hi guys...
This is a simple but a truly simple question....
When trying to create a proxy from our messages interfaces using SPROXY, i'm not able to get the Software component versions that were used from the sld...
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Hi Goncalo,
Check if it helps...
Re: SW component version not visible in SPROXY
Regards
Anand

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a good site for the info www.xml.com or www.oasis.org
xml-spy is a good tool

Question about creating a delay in AS 20

Hey everyone. Let me start off by saying that although I
design in flash often, my knowledge of ActionScript is limited at
best. I am typically a graphic artist, using Adobe CS2 etc.
That said, I have been involved in the updating of a site
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tweened onto the stage. I have created a graphic loader myself,
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What I'm trying to find out is, can I have the script begin
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tmclL.onLoadInit = function() {
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removeMovieClip("thumbLoading");
p1w = myPanel1._width;
p2w = myPanel2._width;
pt = -(((p1w + p2w + thumbPad) - myMask1._width) - maskX *
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tw11.onMotionStopped = function() {
myPanel1.onEnterFrame = scrollUp;
myPanel1.onRollOver = myPanel2.onRollOver = panelOver;
loadImage(images[0]);
I have tried inserting my own "if" statements with little
luck, so if anyone could provide me with some insight, I would be
hugely appreciative.
Thanks!

Yes, you can do that, but it's a little difficult to advise
you the best way with the information provided:
Can you clarify:
Do you want to play the main clip while the thumbs continue
to load... or do you want to preload the thumbs, then play the main
clip, then do the tweening when the main timeline reaches the end?
This way is easy enough.
If you want them to continue to load and then do the tweening
at the end of the main timeline (e..g with a stop() on the last
frame) then its also not complicated, but you need to cover the
situation where the loading of the thumbs may not yet have
completed by that time (perhaps the risk is low if the timeline is
long, but its a sensible thing to address regardless).
Also consider whether you want anything to be hidden (i.e.
_visible=false) at any point - e.g. before the thumbs have finished
loading.
As a starting point I suggest you would probably put the code
after
removeMovieClip("thumbLoading");
and before the last curly bracket (" } ") in a separate
function on the main timeline, e.g.
function tweenThumbs(){
//code here
(its actually not easy to be sure, because I can't visualize
what exactly it does)
what you would do next depends on the answers to the above
questions.

Question about generate key press event to system. help plz.

I am wondering if it is possible to write a java program to generate key press event(not receiving)to the local system so all the other program in the system receive those key press event also. Thanks.

Hi,
:. I don't know what exactly you intend to accomplish. But, I have used the following code in order to simulate keys pressing inside a Java application.
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      EventQueue evtq = Toolkit.getDefaultToolkit().getSystemEventQueue();
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      evtq.postEvent( new KeyEvent(this, KeyEvent.KEY_PRESSED, 0,
                      InputEvent.SHIFT_DOWN_MASK, KeyEvent.VK_TAB,
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      evtq.postEvent( new KeyEvent(this, KeyEvent.KEY_RELEASED, 0,
                      InputEvent.SHIFT_DOWN_MASK, KeyEvent.VK_TAB,
                      KeyEvent.CHAR_UNDEFINED) );:. However, as far as I know to send keys to the whole OS you have to create a 'Hook' as described inside Win32API documentation. That's for Microsoft Windows naturaly.
Cheers.
Roque

Question about DBCA generate script o create RAC database 2 node cluster

Question about creating two node RAC database 11g after installing and configuration 11g clusterware. I've used DBCA to generate script to create a rac database. I've set
environment variable ORACLE_SID=RAC and the creating script creates instance of RAC1 and RAC2. My understanding is that each node will represent a node, however there should only be one database with a name of 'RAC'. Please advise

You are getting your terminology mixed up.
You only have one database. Take a look, there are one set of datafiles on shared storage.
You have 2 instances which are accessing one database.
Database name is RAC. Instance names are RAC1, RAC2, etc, etc.
Also, if you look at the listener configuration and if your tnsnames is setup properly then connecting to RAC will connect you to either one of the instances wheras connecting to RAC1 will connect you to that instance.

A Question about RAW and Previews

I have just recently starting shooting in RAW (mostly for the post production editing abilities - I am an avid amateur photographer bent on learning as much as I can). I set my camera to capture in RAW + L. I don't know why I feel like I want it to capture both the RAW and JPEG file, and thus leads me to my first question: Is it necessary to have the camera capture both the RAW and Large JPEG? I am assuming the answer to be no, as I am sure if after importing the RAW file into Aperture, you could always export a JPEG if you wanted one? So no need to fill up your internal memory (if using managed masters) with the extra JPEG? Is this thinking correct?
Next, if you do import RAW-only files and then want to export certain images, do you have a choice to export the original RAW image? It seems that it only allows you to export a JPEG Original Size. To answer my own question, perhaps you have to export the Master in order to export the full RAW file when exporting? If you want to export a JPEG, you have to export not the Master, but a version of the Master? Is this correct?
Lastly, I wanted to ask a question about Previews. I have my preferences set so that previews have the highest quality with no limits to size. What is the significance of setting it this way? I just assumed that if I wanted to share an image at the highest quality without exporting it, this was the way to go. Is there any validity to this? The reason I ask is that I don't want to have all of these high quality previews taking up internal disk space if I really don't need to. Is there a way to change the preview size once previews are created? Meaning, if you have it set to generate low quality previews, can you change it dynamically to high and vice versa?
I know this is a lot in one post. Thanks for tackling it.
Mac

You can change the quality of the Previews in the Preferences -> Previews tab.
You can regenerate Previews with the Delete and Update Previews under the Images menu.
Regards
TD

A question about arrays (brace yourself this is a long post)

I am working on an assignment that is about generating a random rabbit race by using arrays:
Rabbit Race
Write a program to simulate a race between 4 rabbits (numbered 0 to 3). Initially, all contestants are at the start of the race track (position 0) and must race on a 20 meter track (positions numbered 0 to 19). Each participant has its own technique; some run fast, but are sometimes lazy; while others run slowly but steadily. All rabbits make 1 movement one after the other and the one who arrives at the end of the track before the others, wins the race. Here are the rabbit?s strategies :
At each turn, rabbit 0:
- sleeps (so does not run), 50% of the time;
- runs 4 meters, 30% of the time;
- runs 3 meters, 20% of the time.
At every turn, rabbit 1 runs 2 meters.
At each turn, rabbit 2:
- sleeps, 30% of the time;
- runs 3 meters, 70% of the time.
At each turn, rabbit 3:
- runs 1 meter, 80% of the time;
- runs 10 meters, 20% of the time.
The difficulty in the race is that the track has little slippery hills. So after each movement, rabbits can slide down a hill bringing them either closer to the finishing line, or closer to the starting line. How much they slide and in what direction depends on the slope of the hill at their current position, and how many meters they ran during this turn. Each position (or meter) of the hill has a slope represented by an integer between -3 and +2. If the slope is :
- zero : then, there is no hill at this position and rabbits that reach this position stay there.
- a negative integer : then, there is a declining side of a hill at this position, and all rabbits reaching this position will slide down
to a new position that is closer to the finishing line. The new position will be the current position minus the slope times the number of meters the rabbit ran.
For example, assume a rabbit was in position 10, then ran 5 meters to position 15. If the slope at position 15 is -2, then the new position of the rabbit will be 15 - (-2 x 5) = 25.
- a positive integer : then, there is a rising side of hill at that position, and all rabbits reaching this position will slide down to a new position that is closer to the starting line. The new position will be the current position minus the slope times the number of meters the rabbit ran.
For example, assume a rabbit was in position 10, then ran 5 meters to position 15. If the slope at position 15 is +1, then the new position of the rabbit will be 15 - (+1 x 5) = 10.
Notes on the race:
- Rabbits have 40 turns to reach the finish line. If after 40 turns no one has managed to reach position 19 (or beyond), then the race is cancelled and no winner is declared.
- A rabbit?s position can never go below 0. If, after calculating a rabbit?s position, you arrive at a negative position (ex. position -3), then consider that it is at position 0.
- Consider that any position outside the track (ex. position 0 or position 20) has a slope of 0.
- If a rabbit reaches position 19 (or goes beyond it), it does not immediately win the race. If any other rabbit reaches position 19 (or beyond) during the same turn, then there is a tie.
- If at the same turn, one rabbit reaches position 19, and another reaches a position higher than 19, the second rabbit does not win the race because it went further than the first. They both win.
Ok I know this is long and boring to read but I already completed the majority of the assignment here is my code:
public class Assignment_3 {
 public static void main (String args[])
 int[] position = new int[20];
 int[] slope = new int[20];
 int[] moveRabbit = new int[4];
 moveRabbit[1] = 2;
 int[] currentPosition = new int[4];
 int currentPositionandSlope = 0;
 for (int i=0;i<position.length;i++)
 { position=i;
 System.out.print(position[i] + " ");
 System.out.println();
 for (int i=0;i<(position.length-5);i++)
 { slope[i]=(int) (-3 + Math.random()*5);
 for (int i=0;i<position.length;i++)
 System.out.print(slope[i] + " ");
 System.out.println();
 for (int turn=1;turn<=40;turn++)
 int randomNb0 = (int) (Math.random () * 100);
int randomNb2 = (int) (Math.random () * 100);
int randomNb3 = (int) (Math.random () * 100);
 if (randomNb0<=50)
 moveRabbit[0] = 0;
 else if (randomNb<=30)
 moveRabbit[0] = 4;
 else
 moveRabbit[0] = 3;
 if (randomNb2<=30)
 moveRabbit[2] = 0;
 else
 moveRabbit[2] = 3;
 if (randomNb3<=80)
 moveRabbit[3] = 1;
 else
 moveRabbit[3] = 10;
for (int j=0;j<=3;j++)
System.out.println ("Rabbit" + j + " is at position " + position[currentPosition[j]]);
if (slope[currentPosition[j]+moveRabbit[i]] < 0)
 currentPositionandSlope = position[currentPosition[j]+moveRabbit[j]] - (slope[currentPosition[j]+moveRabbit[j]] * moveRabbit[j]);
else if (slope[currentPosition[j]+moveRabbit[i]] > 0)
 currentPositionandSlope = position[currentPosition[j]+moveRabbit[j]] - (slope[currentPosition[j]+moveRabbit[j]] * moveRabbit[j]);
else
 currentPositionandSlope = (currentPosition[j]+moveRabbit[j]);
System.out.println ("it will move by " + moveRabbit[j] + ", but at position " + (position[currentPosition[j]] + moveRabbit[j]) + " there is a slope of " + slope[currentPosition[j] + moveRabbit[j]]);
System.out.println ("so it will slip by " + (position[currentPositionandSlope] - moveRabbit[j]) + " positions, and end up at position " + position[currentPositionandSlope]);
currentPosition[j] += currentPositionandSlope;
*Ok basically my question is that in the assignment there is an example output where the rabbit is at position 21 but the array is only 20 in length and my program crashes if i go beyond the array length.*
The example output is here:
Welcome to the 4 rabbit race:
position: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
slopes: 0 0 1 -2 1 0 -1 -1 1 -2 1 0 -1 -1 1 0 0 0 0 0
TURN #1
rabbit 0 is at position 0
it will move by 3, but at position 3 there is a slope of -2
so it will slip by 6 positions, and end up at position 9
rabbit 1 is at position 0
it will move by 2, but at position 2 there is a slope of 1
so it will slip by -2 positions, and end up at position 0
rabbit 2 is at position 0
it will move by 3, but at position 3 there is a slope of -2
so it will slip by 6 positions, and end up at position 9
rabbit 3 is at position 0
it will move by 1, but at position 1 there is a slope of 0
so it will slip by 0 positions, and end up at position 1
TURN #2
rabbit 0 is at position 9
it will move by 0, but at position 9 there is a slope of -2
so it will slip by 0 positions, and end up at position 9
rabbit 1 is at position 0
it will move by 2, but at position 2 there is a slope of 1
so it will slip by -2 positions, and end up at position 0
rabbit 2 is at position 9
it will move by 3, but at position 12 there is a slope of -1
so it will slip by 3 positions, and end up at position 15
rabbit 3 is at position 1
it will move by 1, but at position 2 there is a slope of 1
so it will slip by -1 positions, and end up at position 1
TURN #3
rabbit 0 is at position 9
it will move by 3, but at position 12 there is a slope of -1
so it will slip by 3 positions, and end up at position 15
rabbit 1 is at position 0
it will move by 2, but at position 2 there is a slope of 1
so it will slip by -2 positions, and end up at position 0
rabbit 2 is at position 15
it will move by 3, but at position 18 there is a slope of 0
so it will slip by 0 positions, and end up at position 18
rabbit 3 is at position 1
3
it will move by 10, but at position 11 there is a slope of 0
so it will slip by 0 positions, and end up at position 11
TURN #4
rabbit 0 is at position 15
it will move by 0, but at position 15 there is a slope of 0
so it will slip by 0 positions, and end up at position 15
rabbit 1 is at position 0
it will move by 2, but at position 2 there is a slope of 1
so it will slip by -2 positions, and end up at position 0
*rabbit 2 is at position 18*
*it will move by 3, but at position 21 there is a slope of 0*
*so it will slip by 0 positions, and end up at position 21*
*We have a potential winner...*
rabbit 3 is at position 11
it will move by 1, but at position 12 there is a slope of -1
so it will slip by 1 positions, and end up at position 13
So yeah :) that's basically my question (all this text for a simple question like that lol)
If you've managed to get to the end of my post then any guidance on my code would be appreciated too :)

I'm not going to read that long post. As suggested, produce an SSCCE
and paste in the exact, complete error message, and indicate clearly which line causes it.
I did notice this though:
Ok basically my question is that in the assignment there is an example output where the rabbit
is at position 21 but the array is only 20 in length and my program crashes if i go beyond the array length.Either you're misunderstanding something, or the assignment has a typo. If you have an array of length 20, then the indices are 0..19.
Maybe the instructor is calling index 0 position 1--separating Java's array impelementation details from
colloquial 1-based counting, so that the rabbit starts at "position 1" (index 0) goes through "position 20"
(index 19), and when he's done, he's "at postion 21"--i.e., past the finish line, off the end of the course, etc.
This won't correspond to any index in the array, but can still be considered a logical position in the race. "Oh, he's
at position 21. He's done. There's nothing more to do with the array, as it has no slot corresponding to
this 'done' position."
That's pure speculation on my part. You'll have to ask your instructor or figure out for yourself what's going on.

A quick question about WebDynpro SLD and R/3 with concurrent users

Hello ,
I have a very quick question about Webdynpros and SLD connecting to an R/3 system, when you configure a webdynpro to connect to an R/3 system using SLD, you configure a user name and password from the R/3 for the SLD to use. What I would like to know is when I have concurrent users of my webdynpro, how can I know what one user did in R/3 and what another user did? Is there a way for the users of the web dynpro to use their R/3 credentials so SLD can access the R/3? Like dynamically configuring the SLD for each user?
- I would like to avoid leaving their their passwords open in the code ( configuring two variable to get the users username and password and use these variables as JCO username and password )
Thanks Ubergeeks,
Guy

Hi Guy
You will have to use Single Sign On to achieve this. In the destination you have defined to connect to R/3 , there is an option to 'useSSO' instead of userid and password. This will ensure that calls to R/3 will be with the userid that has logged into WAS. You wont need to pass any passwords because a login ticket is generated from WAS and passed on to R/3. The userid is derived from this ticket.
For this to happen you will have to maintain a trust relation ship between R/3 and your WAS ,there is detailed documentation of this in help files. Configuration is very straight forward and is easy to perform
Regards
Pran

Questions about PDF exporting with InDe CS5.5

Hey all,
A couple questions about exporting to PDF from the latest version of InDe.
First, I have noticed that it seems to take a lot longer to get to a PDF. Any suggestions for how to speed up the process? It took 8 minutes or so to generate a low-res PDF (for print) of a 24pp document with a few placed images and vector graphics. Wow, that's a long time to wait, especially for a proof.
Second, the background task... if I get it going on making that 8-minute PDF and then work some more on the document, what exactly is in the PDF? Usually I save before making a PDF or printing. So, is the last version saved what will be in the PDF?
(As an aside, this ability to work on the doc while generating a PDF seems kind of weird. Generally one makes a PDF for proofing, or even for printing, when all the changes have been made and everything is "final". So, I see no benefit to being able to work on my document while it's making a PDF, as I'm probably finished making revisions for the time being. I have to say that I kind of like the progress bar you get when you make an interactive PDF, as you know you can't work on the document when that's on the screen... )
Thanks as always.

First, I have noticed that it seems to take a lot longer to get to a PDF. Any suggestions for how to speed up the process? It took 8 minutes or so to generate a low-res PDF (for print) of a 24pp document with a few placed images and vector graphics. Wow, that's a long time to wait, especially for a proof.
Yes, this is abnormally long (and too long), something is wrong. What's the full version of InDesign you are running, as reported by holding down Cmd or Control and selecting About InDesign?
Second, the background task... if I get it going on making that 8-minute PDF and then work some more on the document, what exactly is in the PDF? Usually I save before making a PDF or printing. So, is the last version saved what will be in the PDF?
Saving is not related. InDesign makes a database snapshot of your document the moment you begin the PDF export, and makes the export relative to that snapshot, regardless of edits you continue to make during the export process, and regardless of saving. Of course saving first is a good idea, for several reasons, not the least of which it sounds like something's fairly seriously wrong with your document or your InDesign installation.
We recommend you trash your preferences and export your document to IDML and see if either of those things changes this 8-minute behavior...err, assuming you're running 7.5.2.318.
(As an aside, this ability to work on the doc while generating a PDF seems kind of weird. Generally one makes a PDF for proofing, or even for printing, when all the changes have been made and everything is "final". So, I see no benefit to being able to work on my document while it's making a PDF, as I'm probably finished making revisions for the time being. I have to say that I kind of like the progress bar you get when you make an interactive PDF, as you know you can't work on the document when that's on the screen... )
Yeah, I think the primary benefit is if you are likely to work on 2 or more files in parallel, so you can finish A and export A and then switch to B. If you'd like a dialog box to pop up when export is done, check out my exportPop script from this post: ANN: automatic dialog after background export (exportPop.jsx.

Question about LRU in a replicated cache

Hi Tangosol,
I have a question about how the LRU eviction policy works in a replicated cache that uses a local cache for its backing map. My cache config looks like this:
<replicated-scheme>
<scheme-name>local-repl-scheme</scheme-name>
<backing-map-scheme>
<local-scheme>
<scheme-ref>base-local-scheme</scheme-ref>
</local-scheme>
</backing-map-scheme>
</replicated-scheme>
<local-scheme>
<scheme-name>base-local-scheme</scheme-name>
<eviction-policy>LRU</eviction-policy>
<high-units>50</high-units>
<low-units>20</low-units>
<expiry-delay/>
<flush-delay/>
</local-scheme>
My test code does the following:
1. Inserts 50 entries into the cache
2. Checks to see that the cache size is 50
3. Inserts 1 additional entry (as I understand it, this should cause the eviction logic to kick-in)
4. Checks the cache size again, expecting it to now be 20
With HYBRID and LFU eviction policies, the above logic works exactly as expected. When I switch to LRU however, the code at step 2 always returns a value significantly less than 50. All 50 inserts appear to complete successfully, so I can only assume that some of the entries have already been evicted by the time I get to step 2.
Any thoughts?
Thanks.
Pete L.
Addendum:
As usual, in attempting to boil this issue down to its essential elements, I left out some details that turned out to be important. The logic that causes the condition to occur looks more like:
1. Loop 2 times:
2. Create named cache instance "TestReplCache"
3. Insert 50 cache entries
4. Verify that cache size == 50
5. Insert 1 additional entry
6. Verify that cache size == 20
7. call cache.release()
8. End Loop
With this logic, the problem occurs on the second pass of the loop. Step 4 reports a cache size of < 50. This happens with LRU, LFU, and HYBRID-- so my initial characterization of this problem is incorrect. The salient details appear to be that I am using the same cache name each pass of the loop and that I am calling release() at the end of the loop. (If I call destroy() instead, all works as expected.)
So... my revised question(s) would be: is this behavior expected? Is calling destroy() my only recourse?
Message was edited by: planeski

Robert,
Attached are my sample code and cache config files. The code is a bit contrived-- it's extracted from a JUnit test case. Typically, we wouldn't re-use the same cache name in this way. What caught my eye however, was the fact that this same test case does not exhibit this behavior when running against a local cache directly (as opposed to a repl cache backed by a local cache.)
Why call release? Well, again, when running this same test case against a local cache, you have to call release or it won't work. I figured the same applied to a repl cache backed by a local cache.
Now that I understand this is more a byproduct of how my unit tests are written and not an issue with LRU eviction (as I originally thought), it's not a big deal-- more of a curiosity than a problem.
Pete L. Attachment: coherence-cache-config.xml (*To use this attachment you will need to rename 545.bin to coherence-cache-config.xml after the download is complete.) Attachment: LruTest.java (*To use this attachment you will need to rename 546.bin to LruTest.java after the download is complete.)

How to generate a delayed retriggerable pulse using only one counter with PXI 6070E card

Hi
I have a problem in generating a retriggerable delayed pulse with a single counter(triggered through a signal at gate) using PXI 6070E card. VI was developed in NI LabVIEW traditional DAQ Ver.7.1. I have used the "delayed pulse generator config" VI and a "Start counter" & "Stop counter" VIs for the purpose. But there is no output seen at the out terminal of counter. So I introduced a "wait" VI and set it to 1 sec. Now the pulse output appears but some pulses are missing mometarily after every 1 sec interval. (any solution for this)
I have gone through a few similar requests in the forum but they suggest either to use two counters or to generate a finite pulse train which does'nt fit my application. Moreover PXI 6070E has only 2 counter timers. I am already using one counter to measure the frequency of a pulse train(signal 1). The application requires to generate a delayed retriggerable pulse for every pulse in signal 1. So I have only one counter left.
Can I measure the frequency of signal 1 by analog means.? so that I can use two counters for pulse generation. (Signal 1 is a TTL signal).
Request some help.
Thanks in Advance
Regards

A finite pulse train (N_Pulses >= 2) does require the use of 2 counters on most of our older hardware including your 6070E. If you're just talking about generating a single retriggerable pulse, you would only need one counter.
Here's an example in Traditional DAQ that shows you how to set a retriggerable pulse generation (it also allows you to adjust the characteristics of the pulse on-the-fly).
If you're writing a new program, you might consider switching to DAQmx as it supports NI's latest hardware and recent OSes should you ever need to upgrade. Traditional NI DAQ is no longer in active development. Here's an example of how to implement a retriggerable pulse generation in DAQmx. You should take note that you can't use the two drivers to simultaneously talk to the same piece of hardware, although you should be able to go back and forth by resetting the Traditional DAQ driver before switching to DAQmx.
Best Regards,
John Passiak

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As I regularly back-up my Domain file from my G5 iMac to a FW external drive that sits on my desk, I have just noticed that the file size is now just over 1 Gb.
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Question about generating a delay :-P

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