Question

Integer.toBinaryString

Asked by: bigbong00

I am converting an integer to a binary string. There is however 1 problem. I use this piece of code throughout my project. The problem occured when I got to the number 6. I need java to show me the four bit binary representation for 6 which is 0110 but java outputs this at 110. Is there anyway around this. Likewise for number 5, 4, 3, 2, 1

Here is my code so you can seee an example. You can see at charAt(3) java gives me the error
Does java have a way around this

if (count == 4 && binaryStringValue.charAt(count) == '1')
    		 {
    		 int binaryPosition = 6;  
    		 ham = Integer.toBinaryString(binaryPosition) 
    		 					if (ham.charAt(0)=='1')
    		 position4 = position4 + 1;
    		  					if (ham.charAt(1)=='1')
    		 position3 = position3 + 1;
    		  					if (ham.charAt(2)=='1')
    		 position2 = position2 + 1;
    		 					if (ham.charAt(3)=='1')
    		 position1 = position1 + 1;
    		 }

                                  
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Asked On
2008-02-19 at 06:15:29ID23174337
Tags

Java

Topics

Java Programming Language

,

Programming Languages

Participating Experts
3
Points
300
Comments
8

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Answers

 

by: Bart_CrPosted on 2008-02-19 at 06:54:06ID: 20928840

A bit ugly, but both ways should work:

public class Test {
    private static final String PADDING = "0000";
 
    public static void main(String[] args) {
        String result = String.format("%04d", Integer.valueOf(Integer.toBinaryString(6)));
        System.out.println(result);
 
        String result2 = Integer.toBinaryString(6);
        result2 = PADDING.substring(result2.length()) + result2;
        System.out.println(result2);
 
    }
}
                                              
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by: bigbong00Posted on 2008-02-19 at 07:03:31ID: 20928922

It is a bit too...................................... ugly. How do I put that into my code?

 

by: florin_ganeaPosted on 2008-02-19 at 07:05:12ID: 20928938

Only padding can get you out of the problem.
While converting, only the relevant number of digits are produced due the number of divides made.

 

by: florin_ganeaPosted on 2008-02-19 at 07:07:50ID: 20928963

I think you should change the toBinaryString to include a padding value as parameter, then it is easy.
I think you (should) use a StringBuffer in building the result.
After the conversion, you only have to insert 0-es at the beginning of the string until padding value is reached.

 

by: bigbong00Posted on 2008-02-19 at 07:15:55ID: 20929064

I'm not a java programmer so I look at my program then I look at the code snippet and they look completely different and I don't understand how I insert what I have been advised into my code

 

by: Bart_CrPosted on 2008-02-19 at 07:16:12ID: 20929070

ham = Integer.toBinaryString(binaryPosition);

becomes

ham = String.format("%04d", Integer.valueOf(Integer.toBinaryString(6)));

 

by: florin_ganeaPosted on 2008-02-19 at 07:39:34ID: 20929357

You can, alternatively, use this method:

String toBinaryString(int number, int padding) {
  StringBuffer sbuff = new StringBuffer(Integer.toBinaryString(number));
  while ( sbuff.length() < padding ) {
     sbuff.insert(0, '0');
  }
  return sbuff.toString();
}

 

by: PaulCaswellPosted on 2008-02-19 at 14:21:29ID: 20933221

or most simply, change:

                 ham = Integer.toBinaryString(binaryPosition)

to

                 ham = Integer.toBinaryString(binaryPosition)
                 ham = ("0000"+ham).substring(ham.length);

Although this is not the neatest solution.

Paul

20120131-EE-VQP-002

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