0

我有： 

public byte[] bytes = new byte[5];     //BitComp Class
public BitSet bits = new BitSet(40);

以及名为 BitComp 的类中的 getter 和 setter。以下类将所有前 8 位设置为 1(byte[0])。之后，它将所有字节转换为 BitSet。现在，当它将第二位设置为 true 并打印它们时。 

import java.util.BitSet;
public class TestBitSet {

    public void testBit(){

        BitComp comp = new BitComp();       

        comp.bytes[0] |= 0xFF;

        comp.setBits(getBitsFromByte(comp.getBytes()));

        System.out.println(toCharArray(comp.getBits()));

        BitSet bs = comp.getBits();
        bs.set(1,true);

        comp.setBits(bs);

        System.out.println(toCharArray(comp.getBits()));    
    }

     private BitSet getBitsFromByte(byte[] barray) 
        {
            BitSet bits=new BitSet();
            if(barray!=null)
            {
                for (int i=0; i<barray.length*8; i++) 
                {
                    if ((barray[barray.length-i/8-1]&(1<<(i%8)))!= 0) 
                    {
                        bits.set(i);
                    }
                }
            }
            return bits;
        }

     public static char[] toCharArray(final BitSet bs)
        {     
           final int length = bs.length();     
           final char[] arr = new char[length];     
           for(int i = 0; i < length; i++)
           {         
             arr[i] = bs.get(i) ? '1' : '0';     
           }  
          return arr; 
        }

    public static void main(String args[]){
        TestBitSet tbs = new TestBitSet();
        tbs.testBit();
    }

}

输出：00000000000000000000000000000000011111111 <- 0th
0th-> 0100000000000000000000000000000011111111

不应该有任何变化，因为 byte[0] 包含前 8 个元素，我使用 BitSet 操作将第二个元素设置为 1。所以 BitSet 是从 LHS 接近的，而 Byte 数组是从 RHS 存储的。如何解决这个问题？getBitsFromByte 方法有问题吗？请建议。谢谢

4

2 回答 2

0

You don't provide the code for BitComp, however it appears to me that you determine how to translate bits into bytes and what order to print the bytes.

Its entirely up to you what order you what things to be set or printed out.

于 2011-06-01T15:38:52.720 回答
0

只需在您的集合中添加一点数学： ((byteNumber*8)+(7-bitNumber))

于 2013-07-16T13:44:36.507 回答