Error in binary String to Integer converter

Question

I have binary String To Integer method implemented in my code. Problem is that I'm getting wrong last two digits of method output results, e.g.:

SHOULD BE RESULT : 11111111100001001000101110000100

RESULT : 11111111100001001000101110000011

There is always problem with last two digits of result.

Any help would be greatly appreciated.

code:

public static int binaryStringToInteger (String binaryString){
char[] digits = binaryString.toCharArray();
int binaryInteger = 0;
int count = 0;
for(int i=digits.length-1;i>=0;i--)
  {
    if(digits[i]=='1') 
         {
          binaryInteger +=(int)Math.pow(2, count);
         }
    count++;
  }
 return binaryInteger;
}

I don't understand one thing, namely you wrote that expected result is 11111111100001001000101110000100, while your method is returning an int. — Suspended
– Suspended, Commented May 22, 2015 at 23:27
@Suspended method is returning int, and result I'm getting and expected one are printed with Integer.toBinaryString(binaryInteger); — Croatia Boy
– Croatia Boy, Commented May 22, 2015 at 23:31
You're not by chance encountering an overflow problem on your last multiplication by 2, are you? — swingMan
– swingMan, Commented May 22, 2015 at 23:40

ppDemo · Accepted Answer · 2015-05-22 23:49:31Z

2

There is nothing wrong with your code except int type can only hold so much. Int holds 4 bytes that is 8bitsx4 = 32 bits with the first one reserved for sign, negative if 1 and positive if 0. Try using long int which holds 64 bits:

public static long binaryStringToInteger (String binaryString){
    char[] digits = binaryString.toCharArray();
    long binaryInteger = 0;
    int count = 0;
    for(int i=digits.length-1;i>=0;i--)
      {
        if(digits[i]=='1') 
             {
              binaryInteger +=(long)Math.pow(2, count);
             }
        count++;
      }
     return binaryInteger;
    }

Also check this: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

answered May 22, 2015 at 23:49

ppDemo

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1 Comment

redge Over a year ago

Great if a log was the desired outcome but I thought we were looking for int so a sign would be more appropriate

DigitalNinja · Accepted Answer · 2015-05-22 23:48:54Z

1

public static long binaryStringToInteger (String binaryString){
char[] digits = binaryString.toCharArray();
long binaryInteger = 0;
long count = 0;
for(int i=digits.length-1;i>=0;i--)
{
    if(digits[i]=='1') 
     {
      binaryInteger +=(long)Math.pow(2, count);
     }
count++;
}
 return binaryInteger;
}

And when converting binaryInteger:

Long.toBinaryString(binaryInteger);

answered May 22, 2015 at 23:48

DigitalNinja

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2 Comments

redge Over a year ago

Some explanation would be helpful

DigitalNinja Over a year ago

I believe it's because of the same reason ppDemo pointed out. He needs long to hold the large value from Math.pow(2, count). Just one less bit in the binary string and his code works fine, that one additional bit takes it over. docs.oracle.com/javase/tutorial/java/nutsandbolts/…

fabian · Accepted Answer · 2015-05-23 00:10:17Z

1

You got conversion issues:

2^31 > 2^31-1 = Integer.MAX_VALUE

Therefore the following section from the language specification describes the conversion from double to int:

The value must be too large (a positive value of large magnitude or positive infinity), and the result [...] is the largest representable value of type int [...].

Hence the last value added is 2^31-1 and not 2^31 which causes the issue.

answered May 23, 2015 at 0:10

fabian

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Comments

Luke Matthews · Accepted Answer · 2015-05-23 00:04:17Z

I hope this test result might help. It shows you whats going on quite explicitly.

binStr = 1 int = 1
binStr = 11 int = 3
binStr = 111 int = 7
binStr = 1111 int = 15
binStr = 11111 int = 31
binStr = 111111 int = 63
binStr = 1111111 int = 127
binStr = 11111111 int = 255
binStr = 111111111 int = 511
binStr = 1111111111 int = 1023
binStr = 11111111111 int = 2047
binStr = 111111111111 int = 4095
binStr = 1111111111111 int = 8191
binStr = 11111111111111 int = 16383
binStr = 111111111111111 int = 32767
binStr = 1111111111111111 int = 65535
binStr = 11111111111111111 int = 131071
binStr = 111111111111111111 int = 262143
binStr = 1111111111111111111 int = 524287
binStr = 11111111111111111111 int = 1048575
binStr = 111111111111111111111 int = 2097151
binStr = 1111111111111111111111 int = 4194303
binStr = 11111111111111111111111 int = 8388607
binStr = 111111111111111111111111 int = 16777215
binStr = 1111111111111111111111111 int = 33554431
binStr = 11111111111111111111111111 int = 67108863
binStr = 111111111111111111111111111 int = 134217727
binStr = 1111111111111111111111111111 int = 268435455
binStr = 11111111111111111111111111111 int = 536870911
binStr = 111111111111111111111111111111 int = 1073741823
binStr = 1111111111111111111111111111111 int = 2147483647
binStr = 11111111111111111111111111111111 int = -2
binStr = 111111111111111111111111111111111 int = 2147483645
binStr = 1111111111111111111111111111111111 int = -4

redge · Accepted Answer · 2015-05-23 06:27:57Z

0

I think that you have a problem with a signed integer value and so should treat the appropriate char (the last I think) a sign bit as follows.

public static int binaryStringToInteger (String binaryString){
    char[] digits = binaryString.toCharArray();
    int binaryInteger = 0;
    int count = 0;
    for(int i=digits.length-2; i>=0;i--)
    {
        if(digits[i]=='1') 
        {
             binaryInteger +=(int)Math.pow(2, count);
        }
        count++;
    }
    if(digits[digits.length-1] =='1') 
    {
        binaryInteger *= -1 ;
    }

    return binaryInteger;
}

Also Integer.toBinaryString () creates an unsigned binary string.

edited May 23, 2015 at 6:27

answered May 23, 2015 at 0:08

redge

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