效率测试-----过期检测（毫秒级）

测试机器处理效率，或者做效率优化的适合：

 1 #include <time.h>
 2 #include <sys/time.h>
 3 void getCurrentTime(char* buf)    
 4 {   
 5    struct timeval tv;  
 6  
 7    struct timezone tz; 
 8  
 9    struct tm         *p;
10 
11    gettimeofday(&tv,&tz);
12     
13    p = localtime(&tv.tv_sec);  
14 
15    sprintf(buf,"time_now:%d-%d-%d %d:%d:%d.%ld
", 1900+p->tm_year, 1+p->tm_mon, p->tm_mday, p->tm_hour, p->tm_min, p->tm_sec, tv.tv_usec/1000);  
16 }

 1 #include <sys/time.h>
 2 #include <time.h>
 3 #define  LOG_TIME
 4 用法：
 5      struct timeval t0,t1;
 6      #ifdef LOG_TIME
 7         gettimeofday(&t0,NULL);
 8      #endif
 9      #ifdef LOG_TIME
10         gettimeofday(&t1,NULL);
11         LOGE("function:%s,,previewSize:%dx%d,,ts_makeup_skin_cleanDt ==%ld", __FUNCTION__,jframeWidth,jframeHeight,
12                 (1000000 * (t1.tv_sec - t0.tv_sec) + t1.tv_usec - t0.tv_usec)/1000);
13      #endif

过期检测Java层：

 1 int checkdate() {
 2         Date now = new Date();
 3         SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");//可以方便地修改日期格式
 4         String hehe = dateFormat.format( now );
 5         System.out.println(hehe);
 6         Calendar c = Calendar.getInstance();//可以对每个时间域单独修改
 7         int year = c.get(Calendar.YEAR);
 8         int month = c.get(Calendar.MONTH) + 1;
 9         Log.e(TAG, "timeout year = " + year + "  month = " + month);
10         if (year>VALID_YEAR) {
11                return 0;
12         }
13         if ((year==VALID_YEAR) && (month>VALID_MONTH)) {
14            return 0;
15         }
16         return 1;
17     }

过期时间检测C++层：

 1 #define VALID_YEAR 2015
 2 #define VALID_MONTH 3
 3 
 4 static int checkDate()
 5 {
 6     struct timeval tv;
 7     struct tm *tm;
 8     gettimeofday(&tv, NULL);
 9     tm = localtime(&tv.tv_sec);
10     int year   = tm->tm_year+1900;
11     int month  = tm->tm_mon+1;
12     if (year>VALID_YEAR) {
13        return 0;
14     }
15     if ((year==VALID_YEAR) && (month>VALID_MONTH)) {
16        return 0;
17     }
18     return 1;
19 }