char time_return[32];
char date_return[32];
time_t timep;
time_t timep_china;
time_t timep_date15;
struct tm *p = NULL ;
struct tm *p_date = NULL;
time(&timep);
timep_china = timep+60*60*8;
timep_date15 = timep_china-60*60*24*15;
p = gmtime(&timep_china);
printf("p %p\n", p);
p_date = gmtime(&timep_date15);
printf("p_date %p\n", p_date);
sprintf(time_return,"%d/%d/%d %d:%d:%d",(1900+p->tm_year),(1+p->tm_mon),p->tm_mday,p->tm_hour,p->tm_min,p->tm_sec);
sprintf(date_return,"%d/%d/%d %d:%d:%d",(1900+p_date->tm_year),(1+p_date->tm_mon),p_date->tm_mday,p_date->tm_hour,p_date->tm_min,p_date->tm_sec);
printf("timep %ld\n",timep);
printf("time_return %s\n",time_return);
printf("date_return %s\n",date_return);
return 0;
p 0x7f2300dd7780
p_date 0x7f2300dd7780
timep 1643282078
time_return 2022/1/12 19:14:38
date_return 2022/1/12 19:14:38
struct tm 是一个内置的静态变量,由gmtime()函数和localtime()函数共享,所以实际上只有一个tm,每次调用的结果会覆盖之前的值。
https://stackoverflow.com/questions/10530804/gmtime-change-two-pointers-at-the-same-time
直接调用一次gmtime(&timep); 然后得到转换的结果以后再进行加减就好了。或者就是每次调用gmtime()以后立马就输出。
您好,我是有问必答小助手,您的问题已经有小伙伴帮您解答,感谢您对有问必答的支持与关注!