先抛出问题：
有一全局变量，一个工作读，一个工作写，有没有问题？
先抛出答案，不倡议这么做。
临界资源：各工作/线程采取互斥的形式，实现共享的资源称作临界资源。属于临界资源的硬件串口打印、显示等,软件有音讯缓冲队列、变量、数组、缓冲区等。多任务/线程间应采取互斥形式，从而实现对这种资源的共享。
多任务/多线程状况下在写模块时，只须要封装进爱护机制即可。常见的爱护机制无关中断、信号量、互斥锁等。
最近做的几个我的项目遇到的一些非凡场景：
1.校时
定义全局变量
typedef struct{ /*应用软件工夫*/ unsigned int year; unsigned char month; unsigned char day; unsigned char hour; /*时 */ unsigned char minute; /*分 */ unsigned char second; /*秒 */ unsigned int millisecond; /*毫秒*/ unsigned int microsecond; /*微秒*/ unsigned int nanosecond; /*纳秒*/ unsigned int u40mic; int delta; /*日历时与本地RTC的差值*/ int delta_rtc; /*同步RTC与本地RTC的差值*/ int delta_year; /*用于操作系统年份保护*/ int timeElapse; /*零碎上电后运行了多少秒*/ char DateValid; /*年月日无效标识,1为无效*/ char TimeValid; /*时分秒无效标记,1为无效*/ char BusVaild; /*总线工夫无效, 1为无效, 用于飞参授时及文件夹创立*/}STRUCT_CORE_TIME;STRUCT_CORE_TIME gst_TimeState;int TIME_Regulate_Time(int year, int month, int date, int hour, int minute, int second){ struct timespec NewSetTime; struct tm DownTime; SYSTEM_TIME_TYPE sys_Time = 0; RETURN_CODE_TYPE retCode; if((year < 2000)||(year > 2255)) return -1; if((month < 1) || (month > 12)) return -2; if((date < 1) || (date>31)) return -3; if((hour > 23) || (minute > 59) || (second > 59)) return -4; /*当超出年限时，获取delta*/ if(year > 2100){ if(((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)){ gst_TimeState.delta_year = year - 2096; year = 2096; }else{ gst_TimeState.delta_year = year - 2100; year = 2100; } }else{ gst_TimeState.delta_year = 0; } GET_TIME(&sys_Time, &retCode); DownTime.tm_hour = hour; DownTime.tm_min = minute; DownTime.tm_sec = second; DownTime.tm_mday = date; DownTime.tm_mon = month - 1; DownTime.tm_year = year - 1900; DownTime.tm_isdst = 0; /*设置为非夏令时*/ NewSetTime.tv_sec = mktime(&DownTime); /*工夫间接用delta保护，不波及零碎工夫*/ gst_TimeState.delta = NewSetTime.tv_sec - sys_Time / Time_1S; /*LOG4C日志零碎工夫更新*/ sd_settimeofday(gst_TimeState.delta); gst_TimeState.TimeValid = 1; gst_TimeState.DateValid = 1; return 0;}void TIME_Gather_Time(){ struct timespec NewSetTime; struct tm DownTime; SYSTEM_TIME_TYPE sys_Time = 0; RETURN_CODE_TYPE retCode; GET_TIME(&sys_Time, &retCode); NewSetTime.tv_sec = sys_Time / Time_1S + gst_TimeState.delta; localtime_r(&NewSetTime.tv_sec, &DownTime); gst_TimeState.year = DownTime.tm_year + 1900 + gst_TimeState.delta_year; gst_TimeState.month = DownTime.tm_mon + 1; gst_TimeState.day = DownTime.tm_mday; gst_TimeState.hour = DownTime.tm_hour; gst_TimeState.minute = DownTime.tm_min; gst_TimeState.second = DownTime.tm_sec; gst_TimeState.millisecond = sys_Time / 1000000 % 1000; gst_TimeState.microsecond = sys_Time / 1000 % 1000; gst_TimeState.nanosecond = sys_Time % 1000; gst_TimeState.u40mic = ((gst_TimeState.hour * 3600 + gst_TimeState.minute * 60 + gst_TimeState.second)*1000 + gst_TimeState.millisecond)*25;}校时线程做如下的周期调用：
if(stBusTime.delta和gst_TimeState.delta相差2秒){ /*校对利用工夫*/ Ret = TIME_Regulate_Time(stBusTime.year + 2000, stBusTime.month, stBusTime.day, stBusTime.hour, stBusTime.minute, stBusTime.second); /*只有VCM校时，就给DLR也校时。有的时候音视频文件创建太早，导致视频头内容有问题*/ TIME_Gather_Time(); /*立即TIME_Gather_Time数据，更新gst_TimeState构造体所有内容*/}另一个线程间接取全局变量gst_TimeState内的年月日时分秒。
这种状况实际上是不加爱护的一个线程读，另一个线程写，然而不加爱护也能应用的起因有两点：
1.外界有卫星相对工夫更新gst_TimeState本地工夫
2.对工夫的精度要求不高
获取卡状态
嵌入式零碎下，工作1进行获取卡状态操作：
static void Msg_04_Process(void * const Handler, unsigned char * const Message, int const LenBytes){ DEVICE_INNER_TYPE *pDevice = (DEVICE_INNER_TYPE*)Handler; /*04音讯解决（周期接管）*/ memcpy(&DLR_2_CPM_04, Message, sizeof(DLR_2_CPM_04)); /*IDR端的定义是相同的*/ pDevice->CardState = !DLR_2_CPM_04.Card_State;}工作2周期获取卡状态，并进行一些操作
int PSSS_DLR_Card_State(void * const Handler){ DEVICE_INNER_TYPE *pDevice = (DEVICE_INNER_TYPE*)Handler; int cardState = 0; cardState = pDevice->CardState; return cardState;}利用场景：
Msg_04_Process获取卡号的周期是2秒，而PSSS_DLR_Card_State大概20ms就要被调用一次。
计划：
1.给pDevice->CardState加信号量爱护
2.PSSS_DLR_Card_State中屡次判断pDevice->CardState的值，例如间断3次是同一个值才示意获取胜利