Java解析字節

Question

我有一個字節b

字節有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

我將如何解析第3位和第4位？

Answer 1

您可以使用BitSet類從字節值中檢索特定的位：

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;
}

通過使用上述方法，您可以獲取字節的BitSet表示形式並獲取特定的位：

byte b = ...; // byte value.
System.out.println(fromByte(b).get(2));  // printing bit #3
System.out.println(fromByte(b).get(3));  // printing bit #4

Answer 2

嘗試

    boolean still = (b & 0xC) == 0x0;
    boolean noStill = (b & 0xC) == 0x4;  
    boolean stationed = (b & 0xC) == 0x8; 
    boolean slow = (b & 0xC) == 0xC;

Answer 3

bitwise AND （ & ）

例：

 myByte & 0x08 --> myByte & 00001000 --> 0 if and only if bit 4 of "myByte" is 0; 0x08 otherwise

Answer 4

如果我理解正確，您希望像這樣解析b[3]和b[4]的位：

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

我會這樣做：

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 */}
}

Answer 5

switch ((b>>3)&3){
  case 0: return still;
  case 1: return noStill;
  case 2: return stationed;
  case 3: return slow
}

Answer 6

在JBBP中，它看起來像

@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);