如何在linux下輸出當前時間

 用localtime可直接分解出年月日時分秒

    QUOTE:

     struct tm *ptm;
long ts;
int y,m,d,h,n,s;

ts = time(NULL);
ptm = localtime(&ts);

y = ptm->tm_year+1900; //年
m = ptm->tm_mon+1; //月
d = ptm->tm_mday; //日
h = ptm->tm_hour; //時
n = ptm->tm_min; //分
s = ptm->tm_sec; //秒

    一個有趣的例子：
    QUOTE:

     #include <stdio.h>
#include <time.h>
#include <unistd.h>

int main() {
while (1) {
time_t sec = time(NULL);
struct tm t = *localtime(&sec);

printf("\x1b[2J"); /* clear screen and home cursor */
printf("\x1b[31;40m"); /* red foreground, black background */
printf("\x1b[11;29H"); /* moves cursor to line 11, column 29 */
printf("+-----^--^-----+\n");
printf("\x1b[12;29H");
printf("|\t%02d:%02d:%02d: |\n", t.tm_hour, t.tm_min, t.tm_sec);
printf("\x1b[13;29H");
printf("+-------V------+\n");
sleep(1);
}

return 0;
}