Shift Register while Indexing Arrays in Formula Node

Hello,
I am having trouble using the formula node while indexing my two arrays. I keep getting "NaN" for my output array. I believe the problem lies within my shift register. My guess is that for the first calculation there is not a PreviousAngle so that is why it is giving me "NaN." For the first index I would like the value of the PreviousAngle to be 0. I am new to LabView and unsure what to do.
Thanks!
Christian Seymour
Solved!
Go to Solution.
Attachments:
Complementary Filter.vi ‏21 KB

The shift register should contain 0 by default since you are using a double [if uninitalized, the shfit register takes the default value of the data type. In this case, the default Double is 0], but you should explicitly initalize the shift register to a default value in most cases.
To do so, right click on the "previous angle" shift register and select "Create constant". This will create a Double value of 0 to fill in the register for the first calculation.
If you need a number other than 0 to start, you can just change this constant.
**Note: This is a good habit to have. Shift registers retain value as long as the VI is in memory. So if you run this more than once, it will have the value from the last execution automatically! This is usually unintended behavior for new LabVIEW users -- however you can sometimes take advantage of this feature in some applications and make it intended behavior
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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Array index problem in formula node

dear friend
I'm trying to use formula node to solve a problem because I make this program using c at the begining.
It runs perfectly in turbo c, but somehow it doesn't work in the formula node.
I try to debug my source code, and I find that the formula node seems can't read the true value in
the array.
Here is the code in the formula node.
int32 i,j,k;
for ( i = 0 ; i < sopa; i++ )
j = 1;
ep[i][0] = pa[i];
ep[i][1] = np[ep[i][0]][0];
while ( ep[i][j-1] != exit )
  for ( k = 0 ; k < sonp ; k++ )
    if ( np[ep[i][j-1]][k] != -1 && (pc[ep[i][j-1]][ep[i][j]] + pc[ep[i][j]][exit] >= pc[ep[i][j-1]][np[ep[i][j-1]][k]] + pc[np[ep[i][j-1]][k]][exit] ) && (pc[ep[i][j-1]][np[ep[i][j-1]][k]] == org[ep[i][j-1]][np[ep[i][j-1]][k]] ))
      ep[i][j] = np[ep[i][j-1]][k];
  j++;
This is the part that I think the problem might be.
if ( np[ep[i][j-1]][k] != -1 && (pc[ep[i][j-1]][ep[i][j]] + pc[ep[i][j]][exit] >= pc[ep[i][j-1]][np[ep[i][j-1]][k]] + pc[np[ep[i][j-1]][k]][exit] ) && (pc[ep[i][j-1]][np[ep[i][j-1]][k]] == org[ep[i][j-1]][np[ep[i][j-1]][k]] ))
ep[i][j] = np[ep[i][j-1]][k];
I state that "np[ep[i][j-1]][k] != -1" but I do get the value -1 in the array "ep".
Could someone tell me why I get -1 in the array?
I hope I can finally do it all by labview not using formula node,but array like "pc[ep[i][j-1]][np[ep[i][j-1]][k]] "
seems very hard to wire in labview. Could someone give me some good advices of doing this?
Thanks for your help

altenbach 已寫:
normanshi wrote:
ep[i][0] = pa[i];                                          (A)
ep[i][1] = np[ep[i][0]][0];                            (B)
...    if ( np[ep[i][j-1]][k] != -1 && ... )          (C)
      ep[i][j] = np[ep[i][j-1]][k];                       (D)
I state that "np[ep[i][j-1]][k] != -1" but I do get the value -1 in the array "ep".
Could someone tell me why I get -1 in the array?
I hope I can finally do it all by labview not using formula node,but array like "pc[ep[i][j-1]][np[ep[i][j-1]][k]] "
seems very hard to wire in labview. Could someone give me some good advices of doing this?
There are many possible scenarios to get a -1 in the ep output. For example if the -1 gets already written in step (A) or (B) at a certain index, and that index does not get overwritten in step (D). You are indexing all over the place! Is there some bounds checking on the array contents? What are typical inputs and array sizes?
I'm sure about that the -1 doesn't get already written in step (A) or (B) because ep[i][0] and ep[i][1] is fine. What really makes me confuse is that sometimes ep array has the value like this.
That means it does overwritten some value in step (D). The -1 should never be written into it because I have already state that np[ep[i][j-1]][k] != -1. Somehow I still get -1,and this is really strange! I'm indexing all over the place becuase I can't figure out a better way to achieve my requirement.
I think I should briefly tell you what I'm trying to do. In this program,I input some values which means the distance from one point to another. Then I use floyd's all pairs shortest-path algorithm to find all the shortest distance. Finally I want to find out the "shortest path". I mean like from point 4 to point 0,the shortest path might be like 4 -> 2 -> 1 -> 0, and what the formula node part do is trying to find out this "shortest path". Maybe you can give me some better advice of doing this, I have been thinking a better way to do it for about two months. I think I'm really not good at this.
I try to do it all by labview ,but again I get a strange problem. This time the ep[i][j-1] doesn't get the correct value. In the first time i=0 j-1=0 ep[i][j-1] = 1,but the second time i = 0 j-1 = 1 ep[i][j-1] = 0. This makes the "ep[i][j-1]!=exit" get the incorrect boolean value. The ep[i][j-1] should be 2,not 0. I try to use highlight to find if I make any mistake, but I can't one.This is the input and I put the labview program in attached file (temp4.vi).
Could you help me find out why this happened?
Thanks for your help altenbach!
由 normanshi 在 08-19-2006 10:06 PM 上編輯的訊息
Attachments:
temp4.vi ‏191 KB
ep.jpg ‏10 KB
input.jpg ‏9 KB

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"Adam Russell" wrote in message
news:ajcc8d$1a266r$[email protected]..
>
> "Bill Atkins" wrote in message
> news:[email protected]..
> > Pass a 3-element integer array, containing values 5, 6, 7 into a
> > formula node as input variable x. Pass variable y out. The formula
> > node contains:
> > int32 y;
> > y = (x[2]-x[1]);
> > My result is 6. If the parentheses are removed, the answer is 1.
> > Apparently adding the parentheses causes the expression to be
> > evaluated as y = x[2-1];.
> > Has this been found/fixed? I have mailed NI separately.
>
> It works correctly in 6.1. I'm at home so I can't t
ry it in 6.02.
>
>

Array in formula node

Hello everyone,
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Thanks again for your comments and your time .
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Code_frequency.vi ‏531 KB
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Formula Node
Evaluates mathematical formulas and expressions similar to C on the block diagram. The following built-in functions are allowed in formulas: abs, acos, acosh, asin, asinh, atan, atan2, atanh, ceil, cos, cosh, cot, csc, exp, expm1, floor, getexp, getman, int, intrz, ln, lnp1, log, log2, max, min, mod, pow, rand, rem, sec, sign, sin, sinc, sinh, sizeOfDim, sqrt, tan, tanh. There are some differences between the parser in the Mathematics VIs and the Formula Node.
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Refer to Creating Formula Nodes and Formula Node Syntax for more information about the Formula Node and the Formula Node syntax.
Note The Formula Node accepts only the period (.) as a decimal separator. The node does not recognize localized decimal separators.
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Martin
Solved!
Go to Solution.

Hello,
does it sound well to you ?
@+
Tinnitus
ps: sorry posts mixed
Message Edité par tinnitus le 12-15-2008 03:05 AM
CLAD / Labview 2011, Win Xp
Mission d'une semaine- à plusieurs mois laissez moi un MP...
RP et Midi-pyrénées .Km+++ si possibilité de télétravail
Kudos always accepted / Les petits clicks jaunes sont toujours appréciés
Don't forget to valid a good answer / pensez à valider une réponse correcte
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inc_node.vi ‏6 KB

Shifter register for array of waveforms

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Attachments:
screencap3.png ‏101 KB

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What is the maximum size of indexed array that can be created during execution of while loop?

Hello,
I am using a ring acquisition to take in images. If i write them as files in the loop, it slows down the acquisition unacceptably. Therefore i am converting image to array and storing the array until the acquisition is complete and then reading the array and converting back to image to save. The program sometimes runs out of memory (after a couple of minutes). Can anyone think of a way where i can store the array/images within the execution (without losing image information), without slowing it down and to avoid this memory problem?
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Attachments:
LL_Ring_in_IMAQ_8_bit_array_then_write_as_png.vi ‏94 KB

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HELP: Run-time array dimension in LabVIEW formula node

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shift (LV7.1).vi ‏19 KB

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Sorry but my English is not good.
I'm trying to execute a bit shift left in the Formula Node, but the shift left is not working as I expected.
uInt32 parametro[5];
float32 valorAmostra;
int16 indiceAmostra;
uInt8 indiceResposta;
int16 controle;
int32 valor;
int8 i;
for(i=0; i<5; i++)
parametro[i]=param[i];
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Solved!
Go to Solution.

Portuguese:
Olá,
Existe um bug no LabVIEW para esta operação. Este bug deverá ser corrigido em versões futuras. Uma solução para esta situação é utilizar um int32 ao invés de um ulnt32.
Qualquer dúvida por favor entre em contato.
Obrigado.
English:
I investigated the problem you are having, and it turns out that it is a bug that has already been reported and documented. The work around for it is to use int32 rather than uInt32.
Thanks.
Luciano Borges
R&D Engineer
Pirelli Brazil

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Array Handling in Formula node.vi ‏8 KB

AGovi wrote:
However I require to use complicated formulas such as
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Shift Register while Indexing Arrays in Formula Node

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