bits.java

一个用java写的地震分析软件(无源码)-used to write a seismic analysis software (without source)

package org.trinet.util;

/**
 * Staic classes for bit and byte manipulation.
 */

public class Bits {

/**
 * Convert arbitrary number of bytes to integer converting the specified
 * subarray of bytes using the platform's default character encoding.
 */
    public static int byteStringToInt (byte[] buff, int start, int len)
    {
	String str = new String(buff, start, len);
	Integer i = Integer.valueOf(new String(buff, start, len));
	return ( i.intValue() );	// Integer -> int
    }

/**
 * Intepret 2 bytes as an integer. The bytes are taken from the byte array buff
 * beginning at array index 'start' (first postion is 0).
 * Uses the formula:
 <tt>
  (short)((a << 8) | (b & 0xff))
 </tt)
 * Note that "masking" (& 0xff) is necessary to preserve the original bit pattern
 * because Java has no unsigned data types (except char)
 */
    public static int byteToInt2 (byte[] buff, int start)
    {
	return (buff[start] << 8) | (buff[start+1] & 0xff);
    }

/**
 * Intepret 4 bytes as an integer. The bytes are taken from the byte array buff
 * beginning at array index 'start' (first postion is 0).
 */

    public static int byteToInt4 (byte[] buff, int i)
    {
	return	     ((buff[i]   & 0xff) << 24) |
		     ((buff[i+1] & 0xff) << 16) |
		     ((buff[i+2] & 0xff) <<  8) |
		      (buff[i+3] & 0xff);
    }

/**
 * Convert byte[] array containing 2-byte ints to int[] (4-bytes)
 * How to convert 2 bytes to a signed short...
 * (short)((a << 8) | (b & 0xff))
 * Note that "masking" (& 0xff) is necessary to preserve the original bit pattern
 * because Java has no unsigned data types (except char)
 */

    public static int[] byteArrayToInt2 (byte[] b, int nbytes)
    {
	int[] in = new int[nbytes/2];
	int idx=0;
	short sht;
	for (int i=0; i<nbytes; i=i+2)
	{
	  in[idx] = (b[i] << 8) | (b[i+1] & 0xff);
	  idx++;
	}
	return (in);
    }

/**
 * Convert byte[] array containing 4-byte ints to int[] (4-bytes)
   32-bit integer
   If the bytes read, in order, are
   b1, b2, b3, and b4, where 0 <= b1, b2, b3, b4 <= 255, then the
   result is equal to:

  (((a & 0xff) << 24) | ((b & 0xff) << 16) |
   ((c & 0xff) <<  8) | (d & 0xff))

 */
    public static int[] byteArrayToInt4 (byte[] b, int nbytes)
    {
	int[] in = new int[nbytes/4];
	int idx=0;
	for (int i=0; i<nbytes; i=i+4)
	{

	   in[idx] = ((b[i]   & 0xff) << 24) |
		     ((b[i+1] & 0xff) << 16) |
		     ((b[i+2] & 0xff) <<  8) |
		      (b[i+3] & 0xff);

	   idx++;
	}
	return (in);
    }

/**
 * See if a bit is set.
 * There must be a native Java way to do this but I can't find it.
 * Bits are numbered starting at 0.
 */
    public static boolean bitSet (byte b, int pos)
    {
	int val = (int) Math.pow(2, pos+1);

	return (boolean) ((b & (byte) val)==val);

    }


}