How to get 0-padded binary representation of an integer in java

Question

for example  for 1  2  128  256 the output can be  16 digits    0000000000000001 0000000000000010 0000000010000000 0000000100000000   I tried  String format   16s   Integer toBinaryString 1      it puts spaces for left-padding                   1    How to put 0s for padding  I couldn t find it in Formatter  Is there another way to do it   P S  this post describes how to format integers with left 0-padding  but it is not for the binary representation

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Starting with Java 11  you can use the repeat      method    0  repeat Integer numberOfLeadingZeros i  - 16    Integer toBinaryString i    Or  if you need 32-bit representation of any integer    0  repeat Integer numberOfLeadingZeros i    0   i   1     Integer toBinaryString i

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There is no binary conversion built into the java util Formatter  I would advise you to either use String replace to replace space character with zeros  as in   String format   16s   Integer toBinaryString 1   replace       0     Or implement your own logic to convert integers to binary representation with added left padding somewhere along the lines given in this so  Or if you really need to pass numbers to format  you can convert your binary representation to BigInteger and then format that with leading zeros  but this is very costly at runtime  as in   String format   016d   new BigInteger Integer toBinaryString 1

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Below will handle proper sizes  public static String binaryString int i        return String format       Integer SIZE    s   Integer toBinaryString i   replace       0       public static String binaryString long i        return String format       Long SIZE    s   Long toBinaryString i   replace       0

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Here a new answer for an old post   To pad a binary value with leading zeros to a specific length  try this   Integer toBinaryString   1  lt  lt  len    val   substring  1     If len   4 and val   1   Integer toBinaryString   1  lt  lt  len    val     returns the string  10001   then    10001  substring  1     discards the very first character  So we obtain what we want     0001    If val is likely to be negative  rather try    Integer toBinaryString   1  lt  lt  len     val  amp    1  lt  lt  len  - 1     substring  1

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I was trying all sorts of method calls that I haven t really used before to make this work  they worked with moderate success  until I thought of something that is so simple it just might work  and it did   I m sure it s been thought of before  not sure if it s any good for long string of binary codes but it works fine for 16Bit strings   Hope it helps     Note second piece of code is improved   String binString   Integer toBinaryString 256     while  binString length    lt  16         pad with 16 0 s         binString    0    binString        Thanks to Will on helping improve this answer to make it work with out a loop  This maybe a little clumsy but it works  please improve and comment back if you can      binString   Integer toBinaryString 256   int length   16 - binString length    char   padArray   new char length   Arrays fill padArray   0    String padString   new String padArray   binString   padString   binString

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for int i 0 i lt n i        for int j str i  length   j lt 4 j      str i   0  concat str i        str i  length   is length of number say 2 in binary is 01 which is length 2 change 4 to desired max length of number  This can be optimized to O n   by using continue

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I would write my own util class with the method like below  public class NumberFormatUtils    public static String longToBinString long val        char   buffer   new char 64       Arrays fill buffer   0        for  int i   0  i  lt  64    i            long mask   1L  lt  lt  i          if   val  amp  mask     mask                buffer 63 - i     1                       return new String buffer      public static void main String    args        long value   0b0000000000000000000000000000000000000000000000000000000000000101L      System out println value       System out println Long toBinaryString value        System out println NumberFormatUtils longToBinString value           Output    5 101 0000000000000000000000000000000000000000000000000000000000000101   The same approach could be applied to any integral types  Pay attention to the type of mask  long mask   1L  lt  lt  i

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A naive solution that work would be  String temp   Integer toBinaryString 5   while  temp length    lt  Integer SIZE  temp    0  temp    pad leading zeros temp   temp substring Integer SIZE - Short SIZE     remove excess   One other method would be  String temp   Integer toBinaryString  m   0x80000000    temp   temp substring Integer SIZE - Short SIZE     This will produce a 16 bit string of the integer 5

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This is an old trick  create a string with 16 0 s then append the trimmed binary string you got from String format   s   Integer toBinaryString 1   and use the right-most 16 characters  lopping off any leading 0 s   Better yet  make a function that lets you specify how long of a binary string you want   Of course there are probably a bazillion other ways to accomplish this including libraries  but I m adding this post to help out a friend     public class BinaryPrinter        public static void main String   args            System out format   d in binary is  s n   1  binaryString 1  4            System out format   d in binary is  s n   128  binaryString 128  8            System out format   d in binary is  s n   256  binaryString 256  16               public static String binaryString  final int number  final int binaryDigits             final String pattern   String format     0 dd   binaryDigits            final String padding   String format  pattern  0            final String response   String format    s s   padding  Integer toBinaryString number              System out format    npattern     s  npadding     s  nresponse     s  n n   pattern  padding  response             return response substring  response length   - binaryDigits

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You can use lib https   github com kssource BitSequence  It accept a number and return bynary string  padded and or grouped     String s   new BitSequence 2  16  toBynaryString ALIGN RIGHT  GROUP CONTINOUSLY      return   0000000000000010    another examples    10  -20  30 - gt 00001010 11101100 00011110 i -10- gt 00000000000000000000000000001010 bi 10- gt 1010 sh 10- gt 00 0000 0000 1010 l 10- gt 00000001 010 by -10- gt 1010 i -10- gt bc- gt 11111111 11111111 11111111 11110110

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try     String format   016d n   Integer parseInt Integer toBinaryString 256       I dont think this is the  correct  way to doing this    but it works

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A simpler version of user3608934 s idea  This is an old trick  create a string with 16 0 s then append the trimmed binary string you got     private String toBinaryString32 int i        String binaryWithOutLeading0   Integer toBinaryString i       return  00000000000000000000000000000000               substring binaryWithOutLeading0 length                  binaryWithOutLeading0

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I think this is a suboptimal solution  but you could do  String format   16s   Integer toBinaryString 1   replace       0

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You can use Apache Commons StringUtils  It offers methods for padding strings   StringUtils leftPad Integer toBinaryString 1   16   0

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I do not know  right  solution but I can suggest you a fast patch   String format   16s   Integer toBinaryString 1   replace       0      I have just tried it and saw that it works fine

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This method converts an int to a String  length bits  Either padded with 0s or with the most significant bits truncated   static String toBitString  int x  int bits        String bitString   Integer toBinaryString x       int size   bitString length        StringBuilder sb   new StringBuilder  bits        if  bits  gt  size            for  int i 0  i lt bits-size  i                 sb append  0            sb append  bitString         else         sb   sb append  bitString substring size-bits  size          return sb toString

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import java util Scanner  public class Q3    public static void main String   args        Scanner scn new Scanner System in       System out println  Enter a number         int num scn nextInt        int numB Integer parseInt Integer toBinaryString num        String strB String format   08d  numB    makes a 8 character code     if num gt  1  amp  amp  num lt  255        System out println strB        else          System out println  Number should be in range between 1 and 255

[java] How to get 0-padded binary representation of an integer in java?

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