我正在研究一个解析从家庭自动化模块获得的二进制数据的DLL.
但我需要一些关于我的代码的建议.
所以我得到一个带有一些字节的消息,在这种情况下每个位表示一定的条件.
在我目前的每个条件都是枚举的代码中,我将枚举放在一个数组中,并检查是否设置了相应的位.
private void ParseZoneConditionFlag1(int Flag1) // Flag1 = Hex represenation of byte
{
Zone_Status_ZoneConditionFlagEnum[] FlagArray = new Zone_Status_ZoneConditionFlagEnum[8];
FlagArray[0] = Zone_Status_ZoneConditionFlagEnum.Faulted;
FlagArray[1] = Zone_Status_ZoneConditionFlagEnum.Tampered;
FlagArray[2] = Zone_Status_ZoneConditionFlagEnum.Trouble;
FlagArray[3] = Zone_Status_ZoneConditionFlagEnum.Bypassed;
FlagArray[4] = Zone_Status_ZoneConditionFlagEnum.Inhibited;
FlagArray[5] = Zone_Status_ZoneConditionFlagEnum.Low_Battery;
FlagArray[6] = Zone_Status_ZoneConditionFlagEnum.Loss_Supervision;
FlagArray[7] = Zone_Status_ZoneConditionFlagEnum.Reserved;
base.CheckBitsSet(FlagArray, Flag1, ZoneConditionFlags_List);
}
private void ParseZoneConditionFlag2(int Flag2)
{
Zone_Status_ZoneConditionFlagEnum[] FlagArray = new Zone_Status_ZoneConditionFlagEnum[8];
FlagArray[0] = Zone_Status_ZoneConditionFlagEnum.Alarm_Memory;
FlagArray[1] = Zone_Status_ZoneConditionFlagEnum.Bypass_Memory;
FlagArray[2] = Zone_Status_ZoneConditionFlagEnum.Reserved;
FlagArray[3] = Zone_Status_ZoneConditionFlagEnum.Reserved;
FlagArray[4] = Zone_Status_ZoneConditionFlagEnum.Reserved;
FlagArray[5] = Zone_Status_ZoneConditionFlagEnum.Reserved;
FlagArray[6] = Zone_Status_ZoneConditionFlagEnum.Reserved;
FlagArray[7] = Zone_Status_ZoneConditionFlagEnum.Reserved;
base.CheckBitsSet(FlagArray, Flag2, ZoneConditionFlags_List);
}
Run Code Online (Sandbox Code Playgroud)
我检查实际比特的方法
protected void CheckBitsSet<T>(T[] ConstantArray, int HexValue, List<T> DestinationList)
{
byte b = (byte) HexValue;
for (int i = 0; i < Mask.Length; i++)
{
if(IsBitSet(b, i))
{
DestinationList.Add(ConstantArray[i]);
}
}
}
public bool IsBitSet(byte b, int pos)
{
return (b & (1 << pos)) != 0;
}
Run Code Online (Sandbox Code Playgroud)
这有效,但我想知道是否有更清洁的方法来做到这一点.
使用清洁器,我的意思是无需每次都向阵列添加正确的枚举.
Mar*_*ell 10
怎么样:
[Flags]
enum MyFlags : short
{
None = 0,
Faulted = 1 << 0,
Tampered = 1 << 1,
Trouble = 1 << 2,
Bypassed = 1 << 3,
Inhibited = 1 << 4,
LowBattery = 1 << 5,
LossOfSupervision = 1 << 6,
AlarmMemory = 1 << 8,
BypassMemory = 1 << 9
}
static bool IsSet(MyFlags value, MyFlags flag)
{
return ((value & flag) == flag);
}
Run Code Online (Sandbox Code Playgroud)
并将值读取为2字节值(short注意字节顺序),然后转换为MyFlags.
要检查任何标志,只需:
MyFlags value = ...
bool isAlarmMemory = IsSet(value, MyFlags.AlarmMemory);
Run Code Online (Sandbox Code Playgroud)
当你谈论复合标志时,它会变得棘手,即
bool memoryProblem = IsSet(value, MyFlags.AlarmMemory | MyFlags.BypassMemory);
Run Code Online (Sandbox Code Playgroud)
因为你需要弄清楚你的意思是"是否设置了这些标志?" vs"所有这些标志都设置好了吗?"
它归结为测试;
return ((value & flag) == flag); // means "are all set"
return ((value & flag) != 0); // means "is any set"
Run Code Online (Sandbox Code Playgroud)
阅读:
// this is just some garbage that I'm pretending is a message from
// your module; I'm assuming the byte numbers in the image are
// zero-based, so the two that we want are: \/\/\/ (the 6,3)
byte[] data = { 12, 63, 113, 0, 13, 123, 14, 6, 3, 14, 15 };
// and I'm assuming "byte 7" and "byte 8" (image) are zero-based;
// MyFlags uses byte 7 *first*, so it is little-endian; we can get that
// via:
short flagsRaw = (short)(data[7] | (data[8] << 8));
MyFlags flags = (MyFlags)flagsRaw;
// flags has value Tampered | Trouble | AlarmMemory | BypassMemory,
// which is what we expect for {6,3}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
2979 次 |
| 最近记录: |