how to define an array of fixed length

[LV_Enthu]

hello, i have a basic application in which i need to define an array of size 19; and add 19 elements to it. 

Once 19 elements have been written to the array, i want to start writing to the array from index 0.

It's like in C language, where-in you define array_size in init, and once the array is full, you overwrite the 1st element and so on.

i hope my question makes sense...

any help is appreciated.

[Bob_Schor]

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You can define an array of size 19 with Initialize Array.  For loops are good ways to index through an array -- learn about them.  If you want to simply replace a numbered element, replace Array Subset can work (the subset if an array of 1 element).  More experienced LabVIEW users might use the In Place Element Array Index/Replace Element, but you should start with the For Loop.  There are good tutorials out there ...

BS

[RavensFan]

Rotate 1D Array might be a useful function for you as well.

[Vinal_G]

Hello LV_Enthu,

As Bob said, there is the two way of initialize array, you can follow any of the method which are displayed in image.

[PBP]

Hi Check out these VIs as per your discription. You can try your own method. 

[GerdW]

Hi PBP,

the OP said to use LV8.0, so your LV2012 VIs aren't accessible…

It's like in C language, where-in you define array_size in init, and once the array is full, you overwrite the 1st element and so on.

Try this:

It defines an array with a certain size and fills it by overwriting the elements. Thanks to QR operation it starts over with 1st element once the array "is full"…

[PBP]

Ohh Sorry,

Please Find LV8.0 version.

[JarleEkanger]

I hope this is in a useable version for you. Basically it covers a solution in which you write values at arbitrary intervals. I used an event structure to make my point, but it could of course be a state in a state machine or a while loop or something...

Like RavensFan suggested, it uses the rotate 1D array to rearrange your array. You take the last element and places it first, then replace it with a new value.