Converting binary string from java back into a signed integer

Question

I am using java's built in Integer.toBinaryString(myInt) to convert to a binary string and then I am converting that 32-bit string into an 8-bit string.

My issue lies in that when converting the number back into a signed-integer I lose the sign.

Example:

My Int = -5.

Binary representation = 11111011.

Converting back to integer: 251.

Some of my code:

//Converts an integer to 8-bit binary.
public static String convertTo8BitBinary(int myNum){
    String intToConv = Integer.toBinaryString(myNum);
    //the number is less than 8-bits
    if(intToConv.length()<8){
        String append="";
        for(int i = 8 - intToConv.length(); i>0;i--){
            append += "0";
        }
        intToConv = append+intToConv;
    //the number is more than 8 bits
    }else {
        intToConv = intToConv.substring(intToConv.length() - 8, intToConv.length());
    }
    return intToConv;
}

//Converts an 8-bit binary string to an integer.
public static int convertToIntegerFromBinary(String b){
    return Integer.parseInt(b,2);
}

Any ideas how I can retain the sign? Does the Integer.parseInt(b,2) not work for signed integers? Is there a radix that does work for signed binary?

How about this question: stackoverflow.com/questions/14012013/… ? — Everv0id
– Everv0id, Commented Mar 2, 2015 at 3:46

HJK · Accepted Answer · 2015-03-02 04:03:43Z

1

While representing singed integers, basic machine architecture would consider the uppermost bit for sign. When the uppermost bit set to '1', then its a negative number, otherwise positive number, more accurate integer instead of number. In you case you would be considering the least 8 bits out of 32 bits. But sign bit present at 32nd position hence loosing the sign

You could do bit AND operation as follows: my_32_bit_number & covert this to int (1000 0000 0000 0000 0000 0000 1111 1111) will give you required 8 bit-number

edited Mar 2, 2015 at 4:03

answered Mar 2, 2015 at 3:48

HJK

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2 Comments

Tukajo Over a year ago

Is there no way to work with the 8th bit being the sign?

HJK Over a year ago

Instead of Integer, make use of Byte (apt for 8-bit integers) and consider only 7 bits. Set the 8th bit (MSB-Most Significant bit) to sign of the previous number i.e, the 32th bit.

Guillaume Fouillet · Accepted Answer · 2015-03-03 10:03:02Z

You should manage the most significant bit which is a marker for the sign.

The code below do exactly what you do, but works on (N-1) bits (for 8bits strings, N = 8), then appends 0 or 1 at the start of the result string in order to keep the sign.

//Converts an integer to n-bit binary.
public static String convertToNBitBinary(int myNum, int nBitSigned){
   int nBitUnsigned = nBitSigned -1;

   if(myNum > Math.pow(2,nBitUnsigned) || myNum <= -Math.pow(2,nBitUnsigned) ){
      return "OutOfBound";
   }

   String intToConv = Integer.toBinaryString(Math.abs(myNum));

   //the number is less than nBitUnsigned
   if(intToConv.length()<nBitUnsigned){
      String append="";
      for(int i = nBitUnsigned - intToConv.length(); i>0;i--){
         append += "0";
      }
      intToConv = append + intToConv;
   }else { //the number is more than 8 bits
      intToConv = intToConv.substring(intToConv.length() - nBitUnsigned);
   }
   intToConv = (myNum <0?"1":"0") + intToConv;
   return intToConv;
}

The second method has to be corrected for the same reason :

//Converts an N-bit binary string to an integer.
public static int convertToIntegerFromBinary(String b){
   String number = b.substring(1);

   return (b.charAt(0) == '0'?1:-1)*Integer.parseInt(number, 2);
}

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