Question

C语言怎么把系统时间输出出来？

Answer 1

#include <stdio.h>
#include <time.h>
void main()
{
time_t timep;
time (&timep);
printf("%s",ctime(&timep));
}
运行结果：

Answer 2

代码如下：（还有其他时间函数类型，有需要再问）#include"stdio.h"
#include"time.h"
int main()
{
time_t t;
time(&t);
printf("%s",ctime(&t));
return 0;
}

Answer 3

使用time函数。
time
Defined in header <time.h>
time_t time( time_t *time );

Returns the current calendar time encoded as a time_t object.

Parameters
time - pointer to a time_t object to store the time in or NULL

Return value
Current calendar time encoded as time_t object on success, (time_t)(-1) on error. If the argument is not NULL, the return value is equal to the value stored in the object pointed to by the argument.

Answer 4

用time函数，不过这个不准

Answer 5

代码如下：【还有其他时间函数类型，在后面补充】#include"stdio.h"
#include"time.h"
int main()
{
time_t t;
time(&t);
printf("%s",ctime(&t));
return 0;
} 补充部分：【以下主要用于对所需时间变量的输出】#include <dos.h>
#include <stdio.h>
int main(void)
{
struct date d;
struct time t;
getdate(&d);
gettime(&t);
printf("The current year is: %d\n",d.da_year); /*此变量单输出年*/
printf("The current month is: %d\n",d.da_mon); /*此变量单输出月*/
printf("The current day is: %d\n",d.da_day); /*此变量单输出日*/
printf("The current time is: %2d:%02d:%02d.%02d\n",t.ti_hour, t.ti_min, t.ti_sec, t.ti_hund); /*时间变量分别输出是时、分、秒及秒以下的单位*/ return 0;
}效果如下 希望回答对你有帮助！