Print an integer in binary format in Java

Question

I have a number and I want to print it in binary  I don t want to do it by writing an algorithm  Is there any built-in function for that in Java

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I think it s the simplest algorithm so far  for those who don t want to use built-in functions    public static String convertNumber int a                    StringBuilder sb new StringBuilder                  sb append a  amp  1                 while   a gt  gt  1     0                        sb append a  amp  1                                  sb append  b0                  return sb reverse   toString          Example    convertNumber 1  --   0b1   convertNumber 5  --   0b101   convertNumber 117  --   0b1110101   How it works  while-loop moves a-number to the right  replacing the last bit with second-to-last  etc   gets the last bit s value and puts it in StringBuilder  repeats until there are no bits left   that s when a 0

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Simply try it  If the scope is only printing the binary values of the given integer value  It can be positive or negative         public static void printBinaryNumbers int n            char   arr   Integer toBinaryString n  toCharArray            StringBuilder sb   new StringBuilder            for  Character c   arr              sb append c                     System out println sb             input  5  Output  101

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Here no need to depend only on binary or any other format    one flexible built in function is available That prints whichever format you want in your program   Integer toString int representation    Integer toString 100 8     prints 144 --octal representation  Integer toString 100 2     prints 1100100 --binary representation  Integer toString 100 16    prints 64 --Hex representation

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Since no answer is accepted  maybe your question was about how to store an integer in an binary-file  java io DataOutputStream might be what you re looking for  https   docs oracle com javase 8 docs api java io DataOutputStream html  DataOutputStream os   new DataOutputStream outputStream   os writeInt 42   os flush    os close

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The question is tricky in java  and probably also in other language    A Integer is a 32-bit signed data type  but Integer toBinaryString   returns a string representation of the integer argument as an unsigned integer in base 2   So  Integer parseInt Integer toBinaryString X  2  can generate an exception  signed vs  unsigned    The safe way is to use Integer toString X 2   this will generate something less elegant   -11110100110  But it works

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Old school       int value   28      for int i   1  j   0  i  lt  256  i   i  lt  lt  1  j            System out println j           value  amp  i   gt  0   1   0

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Enter any decimal number as an input  After that we operations like modulo and division to convert the given input into binary number  Here is the source code of the Java Program to Convert Integer Values into Binary and the bits number of this binary for his decimal number  The Java program is successfully compiled and run on a Windows system  The program output is also shown below       public static void main String   args            Scanner sc   new Scanner System in           int integer           String binary               here we count    or null                                      just String binary   null          System out print  Enter the binary Number              integer   sc nextInt             while integer gt 0                        int x   integer   2              binary   x   binary              integer   integer   2                      System out println  Your binary number is     binary           System out println  your binary length       binary length

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It works with signed and unsigned values used powerful bit manipulation and generates the first zeroes on the left   public static String representDigits int num             int checkBit   1  lt  lt   Integer SIZE   8 - 2          avoid the first digit                 StringBuffer sb   new StringBuffer             if  num  lt  0            checking the first digit             sb append  1              else               sb append  0                       while checkBit    0                          if   num  amp  checkBit     checkBit                   sb append  1                  else                   sb append  0                                         checkBit  gt  gt   1                                 return sb toString

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Solution using 32 bit display mask   public static String toBinaryString int n        StringBuilder res new StringBuilder          res  Integer toBinaryString n   or     int displayMask 1 lt  lt 31      for  int i 1 i lt  32 i             res append  n  amp  displayMask   0  0   1            n n lt  lt 1          if  i 8  0  res append                  return res toString         System out println BitUtil toBinaryString 30      O P  00000000 00000000 00000000 00011110

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System out println Integer toBinaryString 343

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This is the simplest way of printing the internal binary representation of an integer  For Example  If we take n as 17 then the output will be  0000 0000 0000 0000 0000 0000 0001 0001  void bitPattern int n             int mask   1  lt  lt  31          int count   0          while mask    0                if count 4    0                  System out print                    if  mask amp n     0                    System out print  0                   else                  System out print  1                  count                mask   mask  gt  gt  gt  1              System out println

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Simple and pretty easiest solution   public static String intToBinaryString int integer  int numberOfBits         if  numberOfBits  gt  0           To prevent FormatFlagsConversionMismatchException           String nBits   String format       numberOfBits    s           Int to bits conversion                 Integer toBinaryString integer                    replaceAll      0              return nBits       returning the Bits for the given int             return null            if the numberOfBits is not greater than 0  returning null

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There are already good answers posted here for this question  But  this is the way I ve tried myself  and might be the easiest logic based   modulo divide add            int decimalOrBinary   345          StringBuilder builder   new StringBuilder             do               builder append decimalOrBinary   2               decimalOrBinary   decimalOrBinary   2            while  decimalOrBinary  gt  0            System out println builder reverse   toString       prints 101011001

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check out this logic can convert a number to any base  public static void toBase int number  int base        String binary           int temp   number 2 1      for  int j   0  j  lt  temp   j              try               binary         number   base               number    base            catch  Exception e                        for  int j   binary length   - 1  j  gt   0  j--            System out print binary charAt j              OR  StringBuilder binary   new StringBuilder    int n 15  while  n gt 0        if  n amp 1   1           binary append 1        else         binary append 0       n gt  gt  1    System out println binary reverse

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Assuming you mean  built-in    int x   100  System out println Integer toBinaryString x      See Integer documentation    Long has a similar method  BigInteger has an instance method where you can specify the radix

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Try this way   public class Bin     public static void main String   args        System out println toBinary 0x94  8           public static String toBinary int a  int bits        if  --bits  gt  0          return toBinary a gt  gt 1  bits    a amp 0x1   0  0   1        else          return  a amp 0x1   0  0   1            10010100

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Binary representation of given int x with left padded zeros   org apache commons lang3 StringUtils leftPad Integer toBinaryString x   32   0

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public static void main String   args         int i   13      short s   13      byte b   13       System out println  i      String format   32s                Integer toBinaryString i   replaceAll       0          System out println  s      String format   16s                Integer toBinaryString 0xFFFF  amp  s   replaceAll       0          System out println  b      String format   8s                Integer toBinaryString 0xFF  amp  b   replaceAll       0          Output   i  00000000000000000000000000001101 s  0000000000001101 b  00001101

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I needed something to print things out nicely and separate the bits every n-bit  In other words display the leading zeros and show something like this   n   5463 output   0000 0000 0000 0000 0001 0101 0101 0111   So here s what I wrote          Converts an integer to a 32-bit binary string     param number         The number to convert     param groupSize         The number of bits in a group     return         The 32-bit long bit string     public static String intToString int number  int groupSize        StringBuilder result   new StringBuilder         for int i   31  i  gt   0   i--            int mask   1  lt  lt  i          result append  number  amp  mask     0    1     0             if  i   groupSize    0              result append                 result replace result length   - 1  result length              return result toString        Invoke it like this   public static void main String   args        System out println intToString 5463  4

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You can use bit mask  1 lt  lt  k  and do AND operation with number  1  lt  lt  k has one bit at k position   private void printBits int x        for int i   31  i  gt   0  i--            if  x  amp   1  lt  lt  i      0               System out print 1            else               System out print 0                       System out println

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for int i   1  i  lt   256  i                  System out print i           show integer         System out println Integer toBinaryString i            show binary         System out print Integer toOctalString i            show octal         System out print Integer toHexString i            show hex

[java] Print an integer in binary format in Java

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