Binary conversion

when converting a integer to binary using Integer.toBinaryString, the leading zeros are removed. Is there a way to keep the leading zeros in the conversion?
any help would be appreciated!
thanks

when converting a integer to binary using
Integer.toBinaryString, the leading zeros are removed.
Is there a way to keep the leading zeros in the
conversion?if by integer you mean int you never had the leading
zeros in the first place ... other that legosa's way, you could be
a bit sneakier, and do:String thirty2 = Integer.toBinaryString(2);
int length = thirty2.length();
String temp = "00000000000000000000000000000000" + thirty2;
thirty2 = temp.substring(length, 32 + length);
System.out.println(thirty2.length() + ": " + thirty2);

Similar Messages

Binary Conversion of file in Emigall(why)

Can any one tell me what is the purpose of converting the file to binary before loading through EMIGALL?Why is it necessary?I have checked the file On application server in the directory where file after binary conversion is placed.I was expecting it to be in binary format i mean combination of 0's and 1's.But it was not.So can anyone please elaborate.
Thanks

Hi Arya
Very strange question.
There's only one valid file format for the EMIGALL upload files at the moment, so obviously the file has to match that particular format.
Not absolutely sure why we chose this file format instead of a delimited TXT file (it was a very long time ago), two things spring to mind:
Inclusion of "special" characters in the content, that could also be used as column delimiter, without compromising the file layout
Incorrect file layout, like missing or additional columns compared to the EMIGALL customising, easier to identify and more likely to fail
Please note, I don't want to start a discussion here why one format is better than another. Simply accept it, it's not a big deal and move on.
I am certain your file is in binary format, you only use the wrong editor to display the file
Yep
Jürgen

ASCII to binary conversion

Does anyone knows how to convert the ASCII value to binary? For example
ASCII 'U' to 01010101?

Hi,
If I understood you correctly, you want to convert a character eg:'U' to
its ASCII value which is 85 (in binary representation 01010101).
With the following method you can convert the first character of a string to
a one byte long unsigned integer.
method BaseClass.CharToInt(input pStr: Framework.string): Framework.ui1
begin
return *(pointer to ui1)(*(pointer to pointer)(&pStr));
end method;
See also the attached .pex file for some more, usefull little methods...
Regards,
Tamas Deak
-----Message d'origine-----
De: Sing Nyguk Ling [mailto:Thomas.Lingbass.com.my]
Date: dimanche 12 mars 2000 09:49
À: kamranaminyahoo.com
Objet: (forte-users) ASCII to binary conversion
Does anyone knows how to convert the ASCII value to binary? For example
ASCII 'U' to 01010101?
For the archives, go to: http://lists.xpedior.com/forte-users and use
the login: forte and the password: archive. To unsubscribe, send in a new
email the word: 'Unsubscribe' to: forte-users-requestlists.xpedior.com

Binary conversion of a big image leads to "java.lang.OutOfMemoryError"?

Hi,
My program loads a black and white 1 bit depth image of around 20 - 30mb. When I try to convert it to binary image, it gives the following error: java.lang.OutOfMemoryError: Java heap spaceI am converting a bufferedimage to binaryimage. I used the following conversion method:
// This method converts an image to a binary image:
     public static BufferedImage convert (BufferedImage src, int type)
          int w = src.getWidth ();
          int h = src.getHeight ();
          BufferedImage dst = new BufferedImage (w, h, type);
        Graphics2D g = dst.createGraphics ();
        g.drawRenderedImage (src, null);
        g.dispose ();
          return dst;
     }The program was working well before but now it keeps giving the heap memory error which is really weird. Is there any way I can fix this? Can anybody suggest a better way to convert a black and white image to binary data. Thanks.

You could increase maximum memory (-Xmx): [http://java.sun.com/javase/6/docs/technotes/tools/windows/java.html]

JMS Adapter binary conversion error

I am getting an MQ message. The message has / embedded when it is picked by JMS Adapter. Because the message content has /, JMS is stripping the message. Is there anyway to get the whole message into XI from JMS without the message getting stripped.
the error message is below
Error while processing message '3c698d81-983c-4d3c-2272-cb505bbd3e4a'; detailed error description: com.sap.aii.adapter.jms.api.channel.filter.MessageFilterException: Error in converting Binary message 1193200ST20080509110023912134297100239120000DI2SAPDVDIROUTI7080509110023912 EDI00000        INITAREQDI2SAPDVBNSF           EBSTMNT 120886882PNCBANK        I N01                                      Y 01,043000096,FORTE,080422,0830,01,80,80,2/                                       02,RCVR,043000096,1,080422,0830,USD,/                                           03,1019283631,USD,030,28192752,,,060,28192752,,,072,000,,,074,000,,/
Edited by: developer on Jul 1, 2008 9:55 PM

hi folks i have resolved that issue it was mapping change...
Now i'm facing another problem .. I m gettin only header record but i don see me items...
my structure is like this field fixed length
HEADER ( 1 )
OrderDetail(Min - 1 and Max - Unbounded)
Trailer(Min - 1 and Max - Unbounded)
configuration of my JMS adapter :
Transform.Class                              :com.sap.aii.messaging.adapter.Conversion
Transform.ContentType                 :text/xml;charset=utf-8
xml.conversionType                      :StructPlain2XML
xml.processFieldNames                :fromConfiguration
xml.documentName                     : MyDocument
xml.documentNamespace           : http://xi.com/test
xml.recordsetStructure               : Header,1,OrderDetail,,Trailer,
xml.recordsetName                     : Dataset
xml.Header.keyFieldName            :abc
xml.Header.keyFieldValue            :x1
XML xml.keyFieldType                 :CaseSensitiveString
xml.Header.fieldFixedLengths     :1,2,3..
xml.Header.fieldNames                :abc,cde,fgh...
Similarly for order and trailer
Kindly help in this regard.. i was not able to find in search

Help with binary conversion project

hey my assignment was to make a program that would say the binary representation of an int. i got my program working and im pretty sure it works for all negatives and 0. my question is can you guys help me with a way to do the same steps with for( or while( ? it seems like i went the long way doing this project and even tho it works theres probably a much simpler way of going about it. theres no interface or anything we just change the value of N and recompile it and run from command prompt so far in our class. thanks for the help
class binary {
public static void main(String[]args){
     int N = -264;
     int M = N*-1;
     int A;
     int A1;
     int B;
     int B1;
     int C;
     int C1;
     int D;
     int D1;
     int E;
     int E1;
     int F;
     int F1;
     int G;
     int G1;
     int H;
     int H1;
     int I;
     int I1;
     if(N==0) {
          System.out.println("The Binary reresentation of "+N+" is = 0");
     if(N<0) {
          A=M/2;
          A1=M%2;
          B = A/2;
          B1 = A%2;
          C = B/2;
          C1 = B%2;
          D = C/2;
          D1 = C%2;
          E = D/2;
          E1 = D%2;
          F = E/2;
          F1 = E%2;
          G = F/2;
          G1 = F%2;
          H = G/2;
          H1 = G%2;
          I = H/2;
          I1 = H%2;
     else {
          A= N/2;
          A1 = N%2;
          B = A/2;
          B1 = A%2;
          C = B/2;
          C1 = B%2;
          D = C/2;
          D1 = C%2;
          E = D/2;
          E1 = D%2;
          F = E/2;
          F1 = E%2;
          G = F/2;
          G1 = F%2;
          H = G/2;
          H1 = G%2;
          I = H/2;
          I1 = H%2;
     if(A1==1 && N<0){
     B1=B1-1;
     C1=C1-1;
     D1=D1-1;
     E1=E1-1;
     F1=F1-1;
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==1 && N<0){
     C1=C1-1;
     D1=D1-1;
     E1=E1-1;
     F1=F1-1;
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==1 && N<0){
     D1=D1-1;
     E1=E1-1;
     F1=F1-1;
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==0 && D1==1 && N<0){
     E1=E1-1;
     F1=F1-1;
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==0 && D1==0 && E1==1 && N<0){
     F1=F1-1;
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==0 && D1==0 && E1==0 && F1==1 && N<0){
     G1=G1-1;
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==0 && D1==0 && E1==0 && F1==0 && G1==1 && N<0){
     H1=H1-1;
     I1=I1-1;
     if(A1==0 && B1==0 && C1==0 && D1==0 && E1==0 && F1==0 && G1==0 && H1==1 && N<0){
     I1=I1-1;
     if(A1==-1){
     A1=A1*-1;
     if(B1==-1){
     B1=B1*-1;
     if(C1==-1){
     C1=C1*-1;
     if(D1==-1){
     D1=D1*-1;
     if(E1==-1){
     E1=E1*-1;
     if(F1==-1){
     F1=F1*-1;
     if(G1==-1){
     G1=G1*-1;
     if(H1==-1){
     H1=H1*-1;
     if(I1==-1){
     I1=I1*-1;
     if(A==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+A1);
     else if(B==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+B1+A1);
     else if(C==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+C1+B1+A1);
     else if(D==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+D1+C1+B1+A1);
     else if(E==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+E1+D1+C1+B1+A1);
     else if(F==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+F1+E1+D1+C1+B1+A1);
     else if(G==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+G1+F1+E1+D1+C1+B1+A1);
     else if(H==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+H1+G1+F1+E1+D1+C1+B1+A1);
     else if(I==0 && N<0){
          System.out.println("The Binary reresentation of "+N+" is = 1"+I1+H1+G1+F1+E1+D1+C1+B1+A1);
     if(A==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+A1);
     else if(B==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+B1+A1);
     else if(C==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+C1+B1+A1);
     else if(D==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+D1+C1+B1+A1);
     else if(E==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+E1+D1+C1+B1+A1);
     else if(F==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+F1+E1+D1+C1+B1+A1);
     else if(G==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+G1+F1+E1+D1+C1+B1+A1);
     else if(H==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+H1+G1+F1+E1+D1+C1+B1+A1);
     else if(I==0 && N>0){
          System.out.println("The Binary reresentation of "+N+" is = 0"+I1+H1+G1+F1+E1+D1+C1+B1+A1);
}

String intAsBinaryString(int n){
String res = "";
for(int i = 1; i!=0; i *=2){
    res = ((n & i != 0)?"1":"0") + res;
return res;
}Pete is right, in order to understand this you need to know how integers are stored.

Binary Conversion of a File

Hi Guys,
The entire issue is on File Handling.
There is one file say input.dat and another file say output.dat
There are two classes Head and Tail which are Serializable. I am creating objects of those two classes.
I am inserting the first object of Head, then first few bytes from input.dat and then the first object of Tail.
Like this goes on till it reaches to the end of file. Till here everything is fine.
Now the output.dat which is generating, I want to convert it into binary format (e.g. 10110)
How can I go about it? After making it in a binary form, how can I retrive the data ?
Please help me out
Regards
Unmesh

1) the size of the byte array buffer is 4096. Why is it so ?It is an arbitrary value just plucked, almost, out of thin air. It can really be any value that you want, but, since computers a binary machines, a power of two is very desirable (in this case, 2^12). Secondly, in terms of buffer sizes, as the buffer size increases you eventually wind up with a diminishing returns issue (e.g., the larger the buffer, the more memory, more management, more time). To best see this, take a large file and start out with a buffer size of 1 byte and time how long it takes to process the entire file. Do this again but double the buffer size, now 2. It should take roughly half as long to process the entire file. Repeat this process until you see that the processing time comes close to hitting a plateau. Most implementations that utilize buffers tend to use 4K and 8K simply because it just seems to be the best payoff with the smallest price.
2) i couldn't understand this line of the code
for (int mask = 128; mask > 0; mask /= 2)
From where are you getting these numbers 128 and 2 ? why are you dividing mask by 2 ?A byte consists of 8 bits (to be more precise, let us call this an octet). Each bit has a value - basic number system thing. So, for an octet, the bit values are (most significant bit first):
7   | 6 | 5 | 4 | 3 | 2 | 1 | 0          bit
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1          decimal bit value
80 | 40 | 20 | 10 | 08 | 04 | 02 | 01         hex bit valueTo liken this to the decimal, base 10, system, bit 0 is the 1's column, bit 1 is the 10's column, bit 2 is the 100's column and so on. The binary system (base 2) is mathematically the same. It is generalized like this for a given value at a given "column" for any number system:
value := b^i
where:
b is the base number system in question
i is the column in question
So, to find the complete value of any number in any number system:
value := db^i + db^i-1 + db^i-2 + ... + db^1 + db^0
where:
b is the base number system in question
i is the column in question
d is the value of the digit in the column in question
I should also note that jummoMa1's approach of using bit shifting is more efficient.

Decimal to binary conversion script

hi
i m very happy to all of u on ur response
i m really happy with this forum
i need a script to convert the decimal number in to binary
number
83---------->1010011
thx

here are two handlers I picked up from James Newton. I
couldn't find them anywhere on his site (
http://nonlinear.openspark.com)
so I paste them directly here.
credits go to James.
on mMakeRuleMap(anInteger)
-- INPUT: <anInteger> must be an integer
-- OUTPUT: Returns a 32-item list of 0s and 1s corresponding
to the
-- binary representation of anInteger
if anInteger < 0 then
tBinary = mMakeRuleMap(the maxInteger + 1 + anInteger)
tBinary[1] = 1
else
tBinary = []
tBits = 32
tBinary[32] = 0
repeat while anInteger
tBinary[tBits] = anInteger mod 2
anInteger = anInteger / 2
tBits = tBits - 1
end repeat
end if
return tBinary
end mMakeRuleMap
on mBinaryToDecimal(aBitList)
-- INPUT: <aBitList> is a list of 0s and 1s (up to 32
items)
-- OUTPUT: a decimal number corresponding to the binary
number
-- defined by <aBitList>
tBitValue = 1
tDecimal = 0
i = aBitList.count
if i = 32 then
if aBitList[1] = 1 then
-- The number is negative
i = 31
tDecimal = the maxInteger + 1
end if
end if
repeat while i
if aBitList
then
tDecimal = tDecimal + tBitValue
end if
tBitValue = tBitValue * 2
i = i - 1
end repeat
return tDecimal
end mBinaryToDecimal

Decimal to binary conversion

hi...anyone here knows the code to convert decimal to binary.Please help me!!!

Here's the answer in a riddle - to fill in the blank line is to understand the algorithm used - have fun!
class BinaryConverter{
   public static void main(String []args) throws Error{
      try{
         new BinaryConverter(Integer.parseInt(args[0]));
      }catch(ArrayIndexOutOfBoundsException e){
         System.out.println("No argument entered");
      } catch(NumberFormatException e){
         System.out.println("Enter whole numbers only");
   private BinaryConverter(int i){
      System.out.println("The binary number is: ");
      convert(i, 0, (long)Math.pow(2, 31));
   private void convert(int num, int i, long max){
      if(num / max==1) System.out.print("1");
      else System.out.print("0");
      num %= max;
      // there is a line deliberately removed here!
      if(i++ % 8 ==0) System.out.print(" ");
      if(max >0) convert(num, i, max);
}

Floating point to binary conversion

Hi
I need to convert a floating point decimal number to bits.
Eg. 0.000532 to be converted to binary(bits).
How do I do this?

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
decimal number?You have the same bit pattern in both cases. In one it's held in a double and will be interpreted as a floating point number according to the IEEE 754 representation. In the other it's held in a long and will be interpreted according to the two's complement representation. But it's the same bitpattern.
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.

Gif to binary conversion

Hello
Can anybody tell me how to convert an .gif image file into a binary file
Thanks

Well, if you have 24 bits, you have to go up to the next largest data type. In this case, that would be an U32. You can use the Slit Number and then Build Array to make sure the bytes go where you want them to. This is assuming you are using LabVIEW.
There are only two ways to tell somebody thanks: Kudos and Marked Solutions
Unofficial Forum Rules and Guidelines

Binary conversion without BIN2DEC

Hi
Due to the lack of large bits with this function, I am in need of a function that can convert large numbers to binary and vice versa. Does anyone have any ideas?
Thanks in advance.

May you be more precise ?
Number is unable to make a difference between
1234567890123451
and
1234567890123453
It will treat both as
1234567890123450
On the other hand, it is able to convert in BIN every integer with 15 digits
So, I don't understand what you wish to get.
If you wish to decipher integers with more than 15 digits, you must use an other tool or you must work with strings and use an AppleScript to decipher these strings.
In fact the Numbers behaviour is a bit more complicated.
In cell B2 I entered :
=1234567890123450+4
In cell B3 I entered :
=1234567890123450+2
In cell B4
I entered :
=B2=B3
which returns TRUE
In C2 and C3, I entered :
=DEC2BIN(B)
As you may see, with two numbers treated as EQUAL by B4, it returns different results.
And,
more funny, the formula
=C2=C3 inserted in C4 returns TRUE
Yvan KOENIG (VALLAURIS, France) jeudi 15 avril 2010 09:47:12

Binary conversion 3

Okay so I have fought through this program that is supposed to take a binary number and convert it to a integer (decimal number) I got it to work, and everything works great my only question now is...
I want it to display the results say in a dialog box as follows
13
1
0
1
1
I am able to get it to display in the console like this but not in a dialog box
Here is my code, any help would be appreciated
import javax.swing.JOptionPane;
public class binaryconverter {
     public static void main(String[] args) {
          String binum_st;
          int binum , newnum, pickednum1 ,pickednum2 ,pickednum3 ,pickednum4 ,pickednum5 ,pickednum6 =0,pickednum7 ,pickednum8,dec;
          binum_st= JOptionPane.showInputDialog(null, "Enter first number");
          binum = Integer.parseInt(binum_st);
          newnum = binum;
          pickednum8 = newnum % 10;
          dec = pickednum8 * 1;
          newnum = newnum / 10;
          pickednum7 = newnum % 10;
          dec = (pickednum7 * 2) + dec;
          newnum = newnum /10;
          pickednum6 = newnum % 10;
          dec = (pickednum6 * 4) + dec;
          newnum = newnum /10;
          pickednum5 = newnum % 10;
          dec = (pickednum5 * 8) + dec;
          newnum = newnum /10;
          pickednum4 = newnum % 10;
          dec = (pickednum4 * 16) + dec;
          newnum = newnum /10;
          pickednum3 = newnum % 10;
          dec = (pickednum3 * 32) + dec;
          newnum = newnum /10;
          pickednum2 = newnum % 10;
          dec = (pickednum2 * 64) + dec;
          newnum = newnum /10;
          pickednum1 = newnum % 10;
          dec = (pickednum1 * 128 ) + dec;
          System.out.println(dec);
          if (pickednum1 == 1)
               System.out.println(pickednum1);
          if ((pickednum2 == 0) && (pickednum1 ==0))
               else
                    System.out.println(pickednum2);
          if ((pickednum3 == 0 )&&(pickednum2==0))
          else
               System.out.println(pickednum3);
          if ((pickednum4==0 )&& (pickednum3 ==0))
               else
                    System.out.println(pickednum4);
          if ((pickednum5==0 )&& (pickednum4 ==0))
               else
                    System.out.println(pickednum5);
          if ((pickednum6==0 )&& (pickednum5 ==0))
               else
                    System.out.println(pickednum6);
          if ((pickednum7==0 )&& (pickednum6 ==0))
               else
                    System.out.println(pickednum7);
          if (pickednum8!=0)
                    System.out.println(pickednum8);
}

Reposted from
http://forum.java.sun.com/thread.jspa?threadID=5187303
&messageID=9731638#9731638.
Why? given that you got an answer there. You also got
several wrong answers, in fact the thread had perhaps
the highest concentration of wrong answers in a short
thread I've ever seen, but there was certainly one
correct answer.But the question is now different, it is console vs. dialog output.
So I think my answer stands.

Audio to binary conversion

Hello, i want to take audio input from microphone and convert it into binary data (stream of 0s and 1s).. Suggestions are welcomed. I'm using Labview 2011.

Hello GerdW,
i tried based on wht i understood from ur post. but, i'm not getting stream of bits here.. please go through the pic and suggest the changes to be made. I would be glad if u can send a vi file for that.
Thnx in advance...
Attachments:
aj.PNG ‏8 KB

BigInteger to binary conversion ?

How do i convert a BigInteger to binary..
for example if the BigInteger value is really large like
"1234567890123456789"

This snipet should work. If the first bit is a '0' the nuber is positive, if its a '1' its negative.
          int size = myBigInteger.bitLength();
          char[] result = new char[size+1];
          result[0] = (myBigInteger.signum() == -1)? '1' : '0';
          for (int bit = 0, idx = size; bit < size; bit++, idx--) {
               result[idx] = myBigInteger.testBit(bit)? '1' : '0';
          String bits = new String(result);

Binary conversion

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