Generate 3 digits floating point random number

hai....How I want to generate 3 digits floating point random number??..plz help me...
Thanks..

Hi gdah,
use some common math functions to get a randopm number in the range from 1 to 5 (but not including 5, see help for "random number"!)... Then format to a string (if needed) or set properties of numeric indicator...
Hey Mike, your first version looked quite different
Message Edited by GerdW on 05-05-2008 09:39 AM
Best regards,
GerdW
CLAD, using 2009SP1 + LV2011SP1 + LV2014SP1 on WinXP+Win7+cRIO
Kudos are welcome
Attachments:
RND_1-5_3Digits.png ‏1 KB

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Reading IEEE Floating-Point 32bit number

I have a program written in C++ that works with single precision float (IEEE) in Windows. When I try to read data from that program in LabView they seem different. Labview also use IEEE single float representation. The numbers in the files are sine wave with amplitude 1 and 30 samples/cycle.
The first numbers are
0
0.207912
0.406737 etc.
Reading in LV the second number is 3E 54 E6 E2 where the left 3 bytes are mantissa and the right exponent and sign. Reading in LV the second number from the program1 I have CE E6 54 3E which is the mantissa on the right as little endian, but the exponent and sign are CE?! Any hint how to read it correctly in LV?
Attachments:
Prgm1.bin ‏1 KB
SGLLvtest.bin ‏1 KB

Hi,
This works for the second file.
Regards,
Wiebe.
"markstab" wrote in message
news:[email protected]..
> I have a program written in C++ that works with single precision float
> (IEEE) in Windows. When I try to read data from that program in
> LabView they seem different. Labview also use IEEE single float
> representation. The numbers in the files are sine wave with amplitude
> 1 and 30 samples/cycle.
> The first numbers are
> 0
> 0.207912
> 0.406737 etc.
> Reading in LV the second number is 3E 54 E6 E2 where the left 3 bytes
> are mantissa and the right exponent and sign. Reading in LV the second
> number from the program1 I have CE E6 54 3E which is the mantissa on
> the right as little endian
, but the exponent and sign are CE?! Any
> hint how to read it correctly in LV?
[Attachment Decode.vi, see below]
Attachments:
Decode.vi ‏21 KB

Floating point Number & Packed Number

Hai can anyone tell me what is the difference in using floating point & packed Number .
when it will b used ?

<b>Packed numbers</b> - type P
Type P data allows digits after the decimal point. The number of decimal places is generic, and is determined in the program. The value range of type P data depends on its size and the number of digits after the decimal point. The valid size can be any value from 1 to 16 bytes. Two decimal digits are packed into one byte, while the last byte contains one digit and the sign. Up to 14 digits are allowed after the decimal point. The initial value is zero. When working with type P data, it is a good idea to set the program attribute Fixed point arithmetic.Otherwise, type P numbers are treated as integers.
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<b>Floating point numbers</b> - type F
The value range of type F numbers is 1x10*-307 to 1x10*308 for positive and negative numbers, including 0 (zero). The accuracy range is approximately 15 decimals, depending on the floating point arithmetic of the hardware platform. Since type F data is internally converted to a binary system, rounding errors can occur. Although the ABAP processor tries to minimize these effects, you should not use type F data if high accuracy is required. Instead, use type P data.
You use type F fields when you need to cope with very large value ranges and rounding errors are not critical.
Using I and F fields for calculations is quicker than using P fields. Arithmetic operations using I and F fields are very similar to the actual machine code operations, while P fields require more support from the software. Nevertheless, you have to use type P data to meet accuracy or value range requirements.
reward if useful

Floating point to binary conversion

Hi
I need to convert a floating point decimal number to bits.
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Now if I convert that decimal number to bits(in
the usual method of dividing by 2),will that be the
exact binary representation of the floating point
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Note that Long has a toString method which allows you to convert the long to a String. The radix in your case is 2 for binary.

How to generate a random number with at least 6 digits?

Hi all,
All is in the question. :-)
I would like to generate a random number but the function "random
number" in Labview isn't accurate enough.
Do you know an algorithm to generate a random number of 6 or more
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Thanks,
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Hi PF,
It's simple. Use 2 random number functions in the same process. Now you
should have 2 distinct random numbers that have the same accuracy. Let's say
LV gives accuracy up to the thousandth (0.XXX), now if you divide one number
by 1000 and then add to the other number you should then have something like
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strings chop off the 0 and dot then concatinate them together. After all
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pt that many digits?
Good luck
Louis.
P-F wrote in message
news:[email protected]..
> Hi all,
>
> All is in the question. :-)
>
> I would like to generate a random number but the function "random
> number" in Labview isn't accurate enough.
>
> Do you know an algorithm to generate a random number of 6 or more
> digit?
>
> Thanks,
>
> PF

How to restrict the decimal place of a floating point number?

Hi,
Here is my code:
public void TwoDecimal(double u){
     String w = Double.toString(u);
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     System.out.println(c);
     if (c <= 5)
        double a = Double.parseDouble(w);
        System.out.println(a);
     else
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}I want to show a floating point number which has 2 digits and 2 decimal places, e.g. 45.82, 29.67. This number is input by user and passed as a parameter.
For those case like the above sample floating point numbers, it can display the proper value of 'c'. e.g. 45.67 will display 5.
However, when I passed 99999, it will show 7; 9999 will return 6, not 5.
So, if the user does not input the '.', does it append 2 implicit chars to it? i.e. 99999.0 and 9999.0. So, that's why it returned 7 and 6 for the length of the string respectively.
How can I fix it?
and
Does it has better algorithm?
Pls advise.
gogo

When dealing with a known precision, in your case hundredths, it is often a good idea to use an integer type and add in the decimals on printing only. This is often the case in banking systems. Almost all of them use integer types, (read long) in pennies to store monitary values. Ever seen someone type in a value for a credit card machine? For something like $20 they press.. "2" "0" "0" "0" The machine knows the lowest denomonation in a cent, so it knows where to put the decimal place. I suggest you do something like this. It also helps to avoid base 2 round off errors.
-Spinoza

Java random number generator

Hi does java have a random number generator. If it does can you point me to it please. Thank you.

you can use "java.Math.random()" this function returns a value which is greater than or equal
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can produce numbers between 0 & 9.Yes, you can do that. But, as stated before, java.util.Random has methods that are easier to use for random int values than using java.Math.random is for random int values. It all depends what the OP wants. The OP hasn't come back with additional questions, or to say that he found his answer. So, we don't know whether he has his answer (he has several options now!) or not. :)

Generate Random number at VB6

Thank you for reading below scenario:
assume user will input 4 digits number to a text
Input Number : 1119
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1191
1911
9111
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The first thing to do is to rewrite in PHP.
-- CJ

_dynSessConf - How to tweak logic to generate the random number.

Hi,
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Regards,
Saurav

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I also favor this crisp random number generator, which is adapted to Java from the book "Numerical Recipes in C++".
http://www.nr.com/cpp-blurb.html
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How to generate a random number within a specified tolerance and control that tolerance.

I am trying to simulate data without using a DAQ device by using a random number generator and I want to be able to control the "range" of the random number generated though. On the front pannel, the user would be able to input a nominal number, and the random number generator would generate numbers within plus or minus .003 of that nominal number. The data would then be plotted on a graph. Anyone know how to do this? Or if it's even possible?
Just for information sake, the random number is supposed to look like thickness measurements, that's why I want to simulate plus and minus .003 because the data could be thinner or thicker than the nominal.
Thanks in advance!!
Solved!
Go to Solution.

You can create a random number with a gaussian probability profile using two equal distributed (0...1) random numbers and applying the Box Muller transform.
I wrote one many years ago, her it is. (LabVIEW 8.0).
Message Edited by altenbach on 04-12-2010 09:13 PM
LabVIEW Champion . Do more with less code and in less time .
Attachments:
BoxMuller.vi ‏10 KB

Maximum size of a floating-point number

Morning all,
Does anyone know what the maximum size of a floating point number can be under Solaris 8 running on
a V880?
Regards,
Richard Breckon

I'm not sure what you're asking.
The maximum size is 128 bits.
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Is there any way to generate random number in CPO

Requirement : -
> I want to generate a random number from set of 1-9 numbers .is there any way in cpo ?
Thanks
Siva

I created a process that uses 3 steps, all of which happen inside the engine rather than having to span out to a script, so it runs a lot faster.
Technically, it's pseudo-random...
Predefine Output Variable "Random Number" as type "Numeric"
Step 1: Format Date
Format string: fffffff
(that's seven lower-case letters "f")
Original date: [Process.Start Time]
Step 2: Format Date
Format string: ff\0\0\0\0\0
Original date: [Process Start Time]
Step 3: Set Variable
Variable to update: [Process.Variables.Output.Random Number]
New value: ([Workflow.Format Date.Formatted Date]-[Workflow.Format Date (2).Formatted Date])/100000
This returns a basically random number between 0 and 1 (so you can mulitply it by your maximum value) based on the numeric fraction of a second of the start time of the process.

Check Floating Point Number

Hello All,
I am having some trouble checking the value of a field with Key Figure type Number with 8 byte floating point. I want to read that field and populate another field with an X if true. For example if that field is equal to 5,0000000000000000E+07 then i want to mark the other field with an 'X'.
The problem is in my code, how do i read that number in the fltp field, such as the number above. my code reads as follows for the 'X' field.
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Thanks everyone in advance

You don't need to worry about converting the code into standard format or floating, just implement your code as you want and it will automatically take care of the conversion. Basically 5,0000000000000000E+07 = 50,000,000.
thanks.
Wond

Need help with a JavaScript to generate random number

Hi
First things first. I don't know the J of JavaScript.
When I was thinking of some readymade solution to generate a random number in Captivate, I found this on a LinkedIn forum.
1) Add a user variable to your project called "randomNumber". It's value can be zero.
2) Add a Text Caption with "$$randomNumber$$" in it, to check whether your script is working.
3) Add a button to the slide. Give it an On Success Action of Execute JavaScript. Turn off it's "Continue Playing the Project" checkbox.
4) Add the following text in the button's Script_Window:
var objCP = document.Captivate;
var rand = 1 + Math.floor(Math.random() * 10);
function onButtonClick(){
objCP.cpEISetValue('randomNumber', rand);
onButtonClick();
However, this is (as the poster in the forum warned) not working for Captivate 7. Can somebody please help?
I am open for any other methods of getting the same result.
Thanks in advance,
Sreekanth

Hi Sreekanth,
When testing, make sure you are testing from a web server where the web address begins with http or https. When setting the captivate variable "randomNumber" try this instead:
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window.onButtonClick();
Be sure to test this from a web server or else local security will prevent the javascript from executing.
Best,
Jim Leichliter

Generate 3 digits floating point random number

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