Reality Check!

[Sam Moran]

I wrote the function below to accomplish the following:
1. Determine if the MSB (most significant bit) is set
2. find the true decimal value from the binary string
I swap the bits and then add one to the LSB.
So, now my question, was there a better or easier way to do this? I am sure that I took the long way there.

 
/*  The function below is used to get the correct decimal value
    of the 16-bit address    
*/
public static int twoscomplement(String num) {
    StringBuffer b = new StringBuffer(num);
    int len = b.length(), dec = 0;
    // Test to see if the most significant bit is set
    // if so enter FOR and change 1's to 0's and vise-a-verse
    if (b.charAt(0) == '1') {
        for(int j=0; j < len; j++) {
            if (b.charAt(j) == '1') 
                b.setCharAt(j, '0');
            else
                b.setCharAt(j, '1');
            }//end of FOR
   // In this FOR loop we add 1 to the LSB, perform a carry the 1
    char carry = '1';
    for(int j=len-1; j>-1; j--) {
        if (b.charAt(j) == '1' & carry == '1') {
           b.setCharAt(j, '0');
           carry = '1';
        }
        else if (b.charAt(j) == '0' & carry == '1') {
           b.setCharAt(j, carry);
           carry = '0';
        }
    }// end of FOR
    num = b.toString(); // get the decimal value of the binary number
    dec = Integer.parseInt(num, 2) * -1; // make it negative by times -1
  } // end IF most significant bit is set to on, 1
  else
    // the MSB was set 0 and no operation was performed
    // other than get the decimal value of the binary number
    dec = Integer.parseInt(num, 2);
    return dec;
}

[ February 20, 2003: Message edited by: Sam Moran ]
[ February 20, 2003: Message edited by: Sam Moran ]
[ February 20, 2003: Message edited by: Sam Moran ]
[ February 20, 2003: Message edited by: Sam Moran ]

[Peter den Haan]

Does this fit the bill?

public static int twoscomplement(String num) {
  return Short.parseShort(num, 2);
}

- Peter

[Sam Moran]

Peter thank you very much for the Reply! This doesn't work because I get the following runtime error:
Exception in thread "main" java.lang.NumberFormatException
at java.lang.Short.parseShort(Unknown Source)
at hextobin.twoscomplement(hextobin.java:36)
at hextobin.main(hextobin.java:109)
Maybe I didn't explain properly, but the string that I am currently trying to get the correct value for is a 16 bit binary number. I wrote the above function to work on any size binary number. I have only proved with 16 bits, and have not tried 32 bits yet. I may need to assign to a long if the number of bits is to large.
Any thoughts or comments? Also, I did try using the XOR, add one, but I still had to multiply by -1 to get the correct result. See below:

    int stmp = Integer.valueOf(num, 2).intValue();
    int newtmp = stmp ^ Integer.valueOf("1111111111111111", 2).intValue();
    System.out.println("XOR value = " + newtmp);
    newtmp = (newtmp + Integer.valueOf("0000000000000001", 2).intValue()) * -1;
    System.out.println("This is the new = " + newtmp);

The above works, but as I stated above I was trying to come up with a generic way to perform the translation of the correct signed decimal value. I assumed that I took the long way around.
Thank you again.
[ February 20, 2003: Message edited by: Sam Moran ]

[Peter den Haan]

Originally posted by Sam Moran:
Peter thank you very much for the Reply! This doesn't work because I get [a java.lang.NumberFormatException]

You're right, parseShort() wants to see a sign and doesn't handle a full 16 bits number including the sign bit. D'oh. Stupid behaviour. Oh well, then we'll have to do the bit fiddling ourselves:

    public static long twoscomplement(String num) {
        long value = 0;
        for (int i = 64; i >= 1; i--) {
            char bit = num.charAt(Math.max(num.length() - i, 0)) ;
            value = (value << 1) + ((bit == '0') ? 0 : 1);
        }
        return value;
    }

Basically, this code goes through all the character bits in turn and shifts them into the long value. There is just one twist -- the Math.max() call ensures that the value is left-padded with the MSB (sign) bit to 64 bits.
Padding with the sign bit is how you extend signed binary numbers to larger sizes; the JVM internally handles widening integer conversions in exactly the same way.
- Peter

[Sam Moran]

Peter, thank you again! This attempt does not work properly either. You can tell me to go away and tell me that I should drop it. I was just curious if there was a better way to do it!?!?
The output looks like this:
C:\temp\java\work>java hextobin
This is the new = 65524
This is the new = 65531
Here you can see I get the wrong result. It should be -12 and -5 respectively.

public static int twoscomplement(String num) {
    StringBuffer b = new StringBuffer(num);
    int len = b.length(), dec = 0;
// Test to see if the most significant bit is set
// if so enter FOR and change 1's to 0's and vise-a-verse
    if (b.charAt(0) == '1') {
        long value = 0;        
        for (int i = len; i >= 1; i--) {             
            char bit = num.charAt(Math.max(num.length() - i, 0)) ;            
            value = (value << 1) + ((bit == '0') ? 0 : 1);        
        }
    long newtmp = value;        
    System.out.println("This is the new = " + newtmp);
}

I made a minor change to grab the max chars in the string sent in. What do you think? Go away( i am kidding of course)!

[Peter den Haan]

Originally posted by Sam Moran:
[...] This attempt does not work properly either.

*grumble* Well, it worked for me.

        System.out.println(twoscomplement("000000000000101"));
        System.out.println(twoscomplement("111111111111011"));
        System.out.println(twoscomplement("000000000001100"));
        System.out.println(twoscomplement("111111111110100"));

This outputs 5, -5, 12 and -12, respectively. I obviously misunderstood what you're trying to do. What are you trying to stick into the function?
- Peter

[Peter den Haan]

I should really add that if your string starts with anything other than a 1, you shouldn't complain that the function returns a positive number...
- Peter

[Sam Moran]

Peter, thanks. I have Java Version 1.3.1, and the integer output from the code snipet that you sent me gave me the result of:
This is the new = 65524
This is the new = 65531
instead of -12 and -5.
I do not expect to get a minus decimal if the MSB is set to zero. I have an IF statement at the beginning of the function to test for this. In the mean time I did find a different approach:

  
public static int twoscomplement(String num) {
    StringBuffer b = new StringBuffer(num);
    int len = b.length(), dec = 0;
// Test to see if the most significant bit is set
// if so enter FOR and change 1's to 0's and vise-a-verse
    if (b.charAt(0) == '1') {
        int value = 0;        
        value = Integer.valueOf("1111111111111111", 2).intValue() ^ Integer.parseInt(num, 2);
        dec = ~value;
    } // end IF most significant bit is set to on, 1
    else
// the MSB was set 0 and no operation was performed
// other than get the decimal value of the binary number
        dec =  Integer.parseInt(num, 2);
    return dec;
}

[Peter den Haan]

No clue why JDK 1.3 would give a different result -- it is almost as if your for loop only down from 32, or as if the left-shift gets evaluated as an int! Are you sure you copied the code as-is?
The problem (if it is a problem) with your code is that it will work only up to 31 bits -- 63 bits if you switch to Long. My snippet does all of them. By the way,

Integer.valueOf("1111111111111111", 2).intValue()

looks like an elaborate way to write 0xffff
- Peter
[ February 21, 2003: Message edited by: Peter den Haan ]

[Sam Moran]

Peter, Thank you again. I have looked at my O'Reilly book, the Sun website and done a search via Google for an explanation of getting a signed decimal value from a 16 bit string representation of a binary number, which was my immediate need. I could not find anything that served as a striaght forward approach. So, I attempted to write a function that would XOR the bits and then add one to the LSB. This worked and I proved it out. While I was working on it I decided to build into the function a more generic approach by testing the length of the string in hopes that if a 32 bit binary string was sent in, it could handle it.
I posted this version in the Performance section of JavaRanch in hopes that someone would look at it and say, "hey stupid why didn't you do it this way?" I am at a loss as to why when I used your rendition it gives my the large non-signed decimal value. Thank you again for the information.

[Peter den Haan]

Last attempt, and then I'll leave you in peace.

public static short twoscomplement(String num) {
  return (short)Integer.parseInt(num, 2);
}

This will handle 16-bit bitstrings only. The cast to short is essential..

public static int twoscomplement(String num) {
  return (int)Long.parseLong(num, 2);
}

Will do the same for 32-bit strings. You can use a case statement to combine them, if you wish:.

public static int twoscomplement(String num) {
  long value = Long.parseLong(num, 2);
  switch (num.length()) {
    case 8: return (byte)value;
    case 16: return (short)value;
    case 32: return (int)value;
    default:
      throw new IllegalArgumentException("Supports only 8, 16 or 32 bits");
  }
}

Not pretty, but arguably better than what you have now.
- Peter