Converting Signed byte array content to decimal with shifting in java

Question

I had a byte array with hex decimal values and I converted that array to decimal values, using the following code

byte[] content = message.getFieldValue( "_Decoder Message" ).data();
int[] decContent = new int[ content.length ];
int i = 0;
for ( byte b : content )
   decContent[ i++ ] = b & 0xff;

now the array looks like decContent = [01 00 05 00 00 00] ;

The first 2 indexes of this array is converted and saved as int value as below

 int traceIdlow = decContent[ 0 ] & 0xFF;
 int traceIdhigh = decContent[ 1 ] & 0xFF;
 int TraceId = traceIdlow | ( traceIdhigh << 8 );

The last 4 indexes needs to be converted to int as well but I am confused about the shifting whether it should be 8 or 16 .

How can I convert the last 4 indexes to int value so that according to the above example the value becomes

00 00 00 05 = 5 in decimal.

Any ideas?

Thanks

Minor (but important) point: You're not converting to "decimal", you're converting numeric values from byte to integer size. Then you're you're splitting some of those values that are in bit-packed form into individual integers. "Decimal" never enters into it. — Hot Licks
– Hot Licks, Commented Aug 28, 2013 at 15:05

Henry · Accepted Answer · 2013-08-28 14:35:42Z

1

Something like this:

int val = decContent[ 2 ] | (decContent[ 3 ] << 8)
  | (decContent[ 4 ] << 16) | (decContent[ 5 ] << 24);

Note that an integer is always signed in Java. If the value is unsigned and can get larger than Integer.MAX_VALUE you have to put it into a long instead.

A little test:

public static void main(String[] args) throws IOException {
    int[] decContent = new int[] {1,0,5,0,0,0};
    int val = decContent[ 2 ] | (decContent[ 3 ] << 8)
              | (decContent[ 4 ] << 16) | (decContent[ 5 ] << 24);
    System.out.println(val);
}

prints the answer 5.

edited Aug 28, 2013 at 14:35

answered Aug 28, 2013 at 14:29

Henry

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1 Comment

Wearybands Over a year ago

I was expecting 5 and it sis giving a huge number

NormR · Accepted Answer · 2013-08-28 14:48:11Z

1

Are you asking how to change the order of the 4 bytes from: b0 b1 b2 b3 to: b3b2b1b0 in an int variable?

  byte[] b = {0x5, 0x44, (byte)0x7f, 0x1};      // convert to 017f4405
  int val = b[3]<<24 |  b[2]<<16 | b[1]<<8 | b[0];
  System.out.println("val="+Integer.toHexString(val)); // val=17f4405

edited Aug 28, 2013 at 14:48

answered Aug 28, 2013 at 14:29

NormR

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3 Comments

Wearybands Over a year ago

and also convert it to unsigned dec

NormR Over a year ago

Need to add AND of 0xFF to strip sign when byte expanded to int.

Hot Licks Over a year ago

@UmairIqbal - See my note above. You may PRESENT the data as decimal, but you're not converting it into decimal unless you create a character representation of it.

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