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control case structure with bolean

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hello everyone i have some problem with my .vi. in this case i want to makes 3 case with control bolean. in this first case if i click the button of 3 member in case "0" can send the string from bolean to write in the visa, then second  if i click the button of 5 member in case "1" can send the string from bolean to write in the visa and the last  if i click the button of x member in case "2" can send the string from bolean to write in the visa. but for now i have some problem if i click 3 member in case "0" and x member in case "2" why the string to in case is always 0 but before to case the value of string can display. but for button of 5 member in case "1" its normal. what wrong about that ? does this have an effect because of the election Boolean To (0,1)  and Compound Arithmetic. do you have for solution ? thank you for advance

 

mamprat_0-1643637514999.png

 

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Message 1 of 7
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Hi mamprat,

 


@mamprat wrote:

i have some problem with my .vi.

do you have for solution ?


We will give (good) suggestions as soon as you attach any code.

We cannot debug/edit/run images with LabVIEW!

 


@mamprat wrote:

in this first case if i click the button of 3 member in case "0" can send the string from bolean to write in the visa,

if i click the button of 5 member in case "1" can send the string from bolean to write in the visa

if i click the button of x member in case "2" can send the string from bolean to write in the visa.

but for now i have some problem if i click 3 member in case "0" and x member in case "2" why the string to in case is always 0 but before to case the value of string can display. but for button of 5 member in case "1" its normal.


I really don't understand what the problem is… (Some real code might help.)

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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Message 2 of 7
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Solution
Accepted by topic author mamprat

That integer can only be 0 or 1.

 

If one of the Booleans is true, it becomes a 1. Orring 3 1's will still be 1.

 

The only way the integer becomes 0 is if none ogf the Booleans is true.

 

Put the integers in an array, and use Boolean Array To Integer.

 

Or add the Booleans, so the first is added once, the 2nd 2 times, and the 3rd 4 times. This resulTs in a 2th complement of the values.

wiebeCARYA_0-1643639189945.png

 

 

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Message 3 of 7
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sorry sir, i forgot to add my vi. this is my vi and my rules 

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Message 4 of 7
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i follow the concept from this videos, and the result is not problem. but may i try apper the problem sir

https://www.youtube.com/watch?v=8w_yK1PhIKc

mamprat_0-1643639856494.png

 

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Message 5 of 7
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Solution
Accepted by topic author mamprat

@mamprat wrote:

i follow the concept from this videos, and the result is not problem. but may i try apper the problem sir

https://www.youtube.com/watch?v=8w_yK1PhIKc

mamprat_0-1643639856494.png

 


Note that they use a +, not an OR.

 

So, they will get:

F F F -> 0

F F T -> 1

F T F -> 1

F T T -> 2

T F F ->1

T F T -> 2

T T F -> 2

T T T -> 3

 

This, to me, only makes sense if 1, 2 and 3 are clicked on and off in the right order. Or for an example...

 

You'd usually try to handle each Boolean individually, avoiding 2^n cases, if that's what you want.

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Message 6 of 7
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I've found the problem, I'm seeing the wrong symbol in the boolean. Thank you for the advice

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