0

我有一个字节 b

字节有 8 位

bits for single byte

0 = status
1 = locale
2 = AUX
bits (3 & 4) relay
 1. 0 (hence 00) still
 2. 1 (hence 01) noStill
 3. 2 (hence 10) stationed
 4. 3 (hence 11) slow
5 = message;
6 = stuff
7 = moreStuff

我将如何解析第 3 位和第 4 位？

4

6 回答 6

2

您可以使用BitSetclass 从字节值中检索特定位：

public static BitSet fromByte(byte b)
{
    BitSet bits = new BitSet(8);
    for (int i = 0; i < 8; i++)
    {
        bits.set(i, (b & 1) == 1);
        b >>= 1;
    }
    return bits;
}

通过使用上述方法，您可以获得BitSet字节的表示并获取特定位：

byte b = ...; // byte value.
System.out.println(fromByte(b).get(2));  // printing bit #3
System.out.println(fromByte(b).get(3));  // printing bit #4
于 2012-12-27T17:10:21.930 回答
2

尝试

    boolean still = (b & 0xC) == 0x0;
    boolean noStill = (b & 0xC) == 0x4;  
    boolean stationed = (b & 0xC) == 0x8; 
    boolean slow = (b & 0xC) == 0xC;
于 2012-12-27T17:10:46.937 回答
1 
switch ((b>>3)&3){
  case 0: return still;
  case 1: return noStill;
  case 2: return stationed;
  case 3: return slow
}
于 2012-12-27T17:16:28.277 回答
1

bitwise AND( &)

例子：

 myByte & 0x08 --> myByte & 00001000 --> 0 if and only if bit 4 of "myByte" is 0; 0x08 otherwise
于 2012-12-27T17:09:34.550 回答
1

如果我猜对了，您希望这些位可以像这样被解析b[3]：b[4]

00 = still
01 = noStill
10 = stationed
11 = slow

我会这样做：

if(b[3] == 0) { // still or noStill
    if(b[4] == 0) {/* still */}
    if(b[4] == 1) {/* noStill */}
}
if(b[3] == 1) { // stationed or slow
    if(b[4] == 0) {/* stationed */}
    if(b[4] == 1) {/* slow */}
}
于 2012-12-27T17:12:03.883 回答
0

在JBBP中看起来像

@Bin(type = BinType.BIT) class Parsed { byte status; byte locale; byte aux; byte relay; byte message; byte stuff; byte moreStuff;}
final Parsed parsed = JBBPParser.prepare("bit status; bit locale; bit aux; bit:2 relay; bit message; bit stuff; bit moreStuff;").parse(new byte[]{12}).mapTo(Parsed.class);
于 2014-08-13T10:53:51.407 回答