Trouble with Elapsed Time function and shift register

[Canucks.canucks]

This is my VI. I want my while loop stops after a certain amount of time. I created a shift register to see how many times the loop was running. the problem is if i enter for example 1 second, it shows a value . Then if i stop my program and run it again with 1 second this shift register value is different. even when i close my program, reopen ,run and enter 1 second the shift register is different. what am i doing wrong?

[altenbach]

• You might want to initialize the shift register.
• You might want to reset your express VI whenever i=0 on the inner loop
• You might want to slow down your loops. No need to spin millions of times per second.

Start with a few simple LabVIEW tutorials....

[Canucks.canucks]

thank you for your suggestion. I added a wait function for 50 m/s and initialized my shift register. however, the problem still exists. also, I have to every time stop the program and enter a new value for how long control. if i just enter a new value and hit run it only run the loop once

[altenbach]

You only mentioned two out of my three suggestions. What happened to the remaining? Also look at the express dialog.

50 m/s is a speed, not a wait. (Meters per second). What did you add and where?

Whenever you trap the code inside inner loops, a simple state machine architecture is probably a better choice.

[Canucks.canucks]

sorry my apologies. I created this VI and initialized. I don't know how to initialize the loop . is there a tutorial available for state machine with shift register?

[Canucks.canucks]

Sorry I meant 50 mili second. This is my vi

[altenbach]

• You still have not connected the RESET input as instructed in my earlier reply.
• Your VI still uses 100% of a CPU core when not doing anything (i.e. when not in the inner loop).
• There is no need to maximize the front panel and diagram to the screen. Annoying!
• You "elapsed time" indicator has the wrong representation. Also a "relative time" would be more appropriate.
• If you initialize everything to default, you don't need to do the same to other controls/indicators. Seems redundant. (none of this would be needed if you would use a proper state machine. Also, a default time of zero does not seem reasonable.
• Your VI cannot be stopped while it is in the inner loop.

[Canucks.canucks]

unfortunately I don't know how to reset the express VI whenever i=0 on the inner loop or stop my vi when inner loop is running.  can you please kindly tell me how to do it

[altenbach]

You have the iteration terminal of the inner loop, so wire an =0 to it and wire the output to the express VI. Right?

As I said, things would be much simpler without stacked loops. They key word is "state machine".

[Canucks.canucks]

you are talking about the i in while loop? you can't wire anything to it except an indicator [ not constant 0.