我使用枚举作为标志:
[Flags]
public enum Partition_Status_ConditionEnum : long
{
Bypass_code_required = 1 << 0,
Fire_trouble = 1 << 1,
Fire = 1 << 2,
Pulsing_Buzzer = 1 << 3,
TLM_fault_memory = 1 << 4,
Reserved = 1 << 5,
Armed = 1 << 6,
Instant = 1 << 7,
Previous_Alarm = 1 << 8,
Siren_on = 1 << 9,
Steady_siren_on = 1 << 10,
Alarm_memory = 1 << 11,
Tamper = 1 << 12,
Cancel_command_entered = 1 << 13,
Code_entered = …Run Code Online (Sandbox Code Playgroud) 我想取一个字节的前4位和另一位的所有位并将它们相互附加.
这是我需要实现的结果:

这就是我现在拥有的:
private void ParseLocation(int UpperLogicalLocation, int UnderLogicalLocation)
{
int LogicalLocation = UpperLogicalLocation & 0x0F; // Take bit 0-3
LogicalLocation += UnderLogicalLocation;
}
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但这并没有给出正确的结果.
int UpperLogicalLocation_Offset = 0x51;
int UnderLogicalLocation = 0x23;
int LogicalLocation = UpperLogicalLocation & 0x0F; // Take bit 0-3
LogicalLocation += UnderLogicalLocation;
Console.Write(LogicalLocation);
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这应该给出0x51(0101 0001)+ 0x23(00100011),
所以我想要实现的结果是0001 + 00100011 = 000100100011(0x123)
我在我的应用程序中包含了一个StreetViewPanoramaFragment,但我还想显示在google地图应用中看到的街景视图的预览.

我怎样才能做到这一点?
我有以下字符串:
<stx>1<rs>aaaa<rs>bbbb<rs>cccc<etx>
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我怎么能以一种在数组中得到1,aaaa,bbbb,cccc的方式拆分它?
我尝试分裂,<rs>但后来我进入<stx>1,aaaa,bbbb,cccc了我的阵列.
那我怎么摆脱它<stx>呢?
stx,rs并且etx是炭灰BTW.
谢谢
我想创建一个日历对象,并将其设置为该年的某一年和一周.
Calendar calendar = Calendar.getInstance();
calendar.set(Calendar.WEEK_OF_YEAR, weekOfYear); // 1
calendar.set(Calendar.YEAR, year); // 2016
setWeekChecked(calendar);
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当我将它传递给setWeekChecked方法时,这是日历对象的toString:
java.util.GregorianCalendar中[时间= ?, areFieldsSet =假,宽大=真,区=美国/纽约,Firstdayofweek可= 1,minimalDaysInFirstWeek = 1,ERA = 1,YEAR = 2016,MONTH = 0,WEEK_OF_YEAR = 1,WEEK_OF_MONTH = 2,DAY_OF_MONTH = 7,DAY_OF_YEAR = 7,DAY_OF_WEEK = 5,DAY_OF_WEEK_IN_MONTH = 1,AM_PM = 0,HOUR = 5,HOUR_OF_DAY = 5,MINUTE = 25,SECOND = 43,微差= 219,ZONE_OFFSET = -18000000,DST_OFFSET = 0 ]
在setWeekChecked方法中:
public void setWeekChecked(final Calendar cal) {
final int targetWeek = cal.get(Calendar.WEEK_OF_YEAR); // Returns 1
final int targetYear = cal.get(Calendar.YEAR); // Returns 2015??
} …Run Code Online (Sandbox Code Playgroud)