How to convert a Binary String to a base 10 integer in Java

Question

I have an array of Strings that represent Binary numbers  without leading zeroes  that I want to convert to their corresponding base 10 numbers  Consider   binary 1011 becomes integer 11 binary 1001 becomes integer 9 binary   11 becomes integer 3   etc     What s the best way to proceed  I ve been exploring java lang number   without finding a direct conversion method  Integer parseInt b  yields an integer EQUAL to the String   e g   1001 becomes 1 001 instead of 9   and does not seem to include a parameter for an output base  toBinaryString does the conversion the wrong direction  I suspect I ll need to do a multistep conversion  but can t seem to find the right combination of methods or subclasses  I m also not sure the extent to which leading zeros or lack thereof will be an issue  Anyone have any good directions to point me

User · Answer

static int binaryToInt  String binary       char   cA   binary toCharArray        int result   0      for  int i   cA length-1 i gt  0 i--             111   length   3  i   2  2  3-3    2  3-2                                0           1           if cA i    1   result  Math pow 2  cA length-i-1             return result

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You need to specify the radix  There s an overload of Integer parseInt   which allows you to   int foo   Integer parseInt  1001   2

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Fixed version of java s Integer parseInt text  to work with negative numbers   public static int parseInt String binary        if  binary length    lt  Integer SIZE  return Integer parseInt binary  2        int result   0      byte   bytes   binary getBytes         for  int i   0  i  lt  bytes length  i              if  bytes i     49                result   result    1  lt  lt   bytes length - 1 - i                         return result

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public Integer binaryToInteger String binary       char   numbers   binary toCharArray        Integer result   0      int count   0      for int i numbers length-1 i gt  0 i--            if numbers i    1  result   int Math pow 2  count            count              return result      I guess I m even more bored  Modified Hassan s answer to function correctly

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int foo   Integer parseInt  1001   2     works just fine if you are dealing with positive numbers but if you need to deal with signed numbers you may need to sign extend your string then convert to an Int  public class bit fun       public static void main String   args            int x   int Long parseLong  FFFFFFFF   16           System out println  x     x                   System out println signExtend  1             x   int Long parseLong signExtend  1    2           System out println  x     x            System out println signExtend  0             x   int Long parseLong signExtend  0    2           System out println  x     x            System out println signExtend  1000             x   int Long parseLong signExtend  1000    2           System out println  x     x            System out println signExtend  01000             x   int Long parseLong signExtend  01000    2           System out println  x     x              private static String signExtend String str             TODO add bounds checking         int n 32-str length            char   sign ext   new char n           Arrays fill sign ext  str charAt 0             return new String sign ext  str           output  x  -1 11111111111111111111111111111111 x  -1 00000000000000000000000000000000 x  0 11111111111111111111111111111000 x  -8 00000000000000000000000000001000 x  8    I hope that helps

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If you re worried about performance  Integer parseInt   and Math pow   are too expensive  You can use bit manipulation to do the same thing twice as fast  based on my experience    final int num   87  String biStr   Integer toBinaryString num    System out println   Input Number      num     toBinary    biStr   int dec   binaryStringToDecimal biStr   System out println  Output Number      dec     toBinary   Integer toBinaryString dec      Where  int binaryStringToDecimal String biString     int n   biString length            int decimal   0    for  int d   0  d  lt  n  d            append a bit 0  i e  shift left       decimal   decimal  lt  lt  1          if biStr d  is 1  flip last added bit 0 to 1      if  biString charAt d      1          decimal   decimal   1     e g  dec   110    00 1   111             return decimal      Output    Input Number  87 toBinary 1010111 Output Number  87 toBinary 1010111

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Now you want to do from binary string to Decimal but Afterword  You might be needed contrary method  It s down below  public static String decimalToBinaryString int value        String str    quot  quot       while value  gt  0            if value   2    1                str    quot 1 quot  str            else               str    quot 0 quot  str                    value    2            return str

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This might work   public int binaryToInteger String binary        char   numbers   binary toCharArray        int result   0      for int i numbers length - 1  i gt  0  i--          if numbers i    1               result    Math pow 2   numbers length-i - 1        return result

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For me I got NumberFormatException when trying to deal with the negative numbers  I used the following for the negative and positive numbers    System out println Integer parseUnsignedInt  11111111111111111111111111110111   2           Output   -9

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I love loops  Yay   String myString    1001001     73   While loop with accumulator  left to right  l doesn t change    int n   0      j   -1      l   myString length    while    j  lt  l  n    n  lt  lt  1     myString charAt j      0    0   1   return n    Right to left with 2 loop vars  inspired by Convert boolean to int in Java  absolutely horrible    int n   0      j   myString length      i   1  while  j--    0  n -   i   i  lt  lt  1    new Boolean myString charAt j      0   compareTo true   return n  gt  gt  1    A somewhat more reasonable implementation   int n   0      j   myString length        i   1  while  j--    0  n     i   i  lt  lt  1     myString charAt j      0    0   1   return n  gt  gt  1    A readable version  p  int n   0  for  int j   0  j  lt  myString length    j          n    2      n    myString charAt j      0    0   1    return n

[java] How to convert a Binary String to a base 10 integer in Java

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