Question

• Im trying to convert an integer of size 4 to byte array but its
  giving me strange value.

   byte[] bytes1 = my_int_to_bb_le(196,4);
     public static  byte[] my_int_to_bb_le(int myInteger,int length)
     {
       return ByteBuffer.allocate(length).order(ByteOrder.LITTLE_ENDIAN).putInt(myInteger).array();
     }     
  

• output :
  -60 0 0 0

• required output as : 196 0 0 0

  Tried the alternate this code but not working as its giving the same
  output

  public static byte[] convertIntByteArray(int intval,int len)     
  {       
     byte[] intarray = new byte[len];          
  
     for(int i=0;i<len;i++)        
     {
         if(i==0)
         {
             intarray[i]=(byte)intval;
         }
         else
             intarray[i]=0;        
      } 
      return intarray;     
  }
  

• output : -60 0 0 0

• required output as : 196 0 0 0

  Thanks

Answer 1

That is because, a (signed) byte is between -128 to 127

public class TestNumber{
  public static void main(String[] args){
    byte b1 = -60;
    int b2 = b1 & 0xFF;
    System.out.println("Without casting "+ b1);
    System.out.println("With casting "+b2);
    System.out.println("Bitwise Trickery " + (b1 & 0xFF));
  }
}

If you want to look at how bits are arranged in the memory for each number, just use the calc.exe supplied with all Windows OS, change the view to Programmer and then, just enter a number and see its bitwise values. That’s what I use all the time for bitwise operations 🙂