函数返回一个包含等价时间的结构

编写下列函数，假定time结构包含三个成员：hours、minutes和seconds（都是 int 类型）。\nstruct time split_time(long total_seconds);\ntotal_seconds是从午夜开始的秒数。函数返回一个包含等价时间的结构，等价的时间用小时\n（023）、分钟（059）和秒（0~59）表示。

这个要怎么改，怎么能够运行出结果呀？刚学这里不太懂😭

一种写法：

#include <stdio.h>
#include <stdlib.h>
struct Time{
    int hours;
    int minutes;
    int seconds;
};
struct Time* split_time(long total_seconds)
{
    struct Time *t = (struct Time*)malloc(sizeof(struct Time));
    t->hours = total_seconds / 3600;
    total_seconds %= 3600;
    t->minutes = total_seconds / 60;
    total_seconds %= 60;
    t->seconds = total_seconds;
    return t;
}

另一种写法：

#include <stdio.h>
#include <stdlib.h>
struct Time{
    int hours;
    int minutes;
    int seconds;
}t = {0,0,0};
struct Time split_time(long total_seconds)
{
    t.hours = total_seconds / 3600;
    total_seconds %= 3600;
    t.minutes = total_seconds / 60;
    total_seconds %= 60;
    t.seconds = total_seconds;
    return t;
}

两个函数调用，供参考：

#include <stdio.h>
#include <stdlib.h>
struct Time{
    int hours;
    int minutes;
    int seconds;
}t = {0,0,0};
struct Time split_time(long total_seconds)
{
    t.hours = total_seconds / 3600;
    total_seconds %= 3600;
    t.minutes = total_seconds / 60;
    total_seconds %= 60;
    t.seconds = total_seconds;
    return t;
}
struct Time* split_time1(long total_seconds)
{
    struct Time *t = (struct Time*)malloc(sizeof(struct Time));
    t->hours = total_seconds / 3600;
    total_seconds %= 3600;
    t->minutes = total_seconds / 60;
    total_seconds %= 60;
    t->seconds = total_seconds;
    return t;
}
int main()
{
    struct Time start, *p_start;
    start = split_time(156800);  // 方式一
    printf("%d:%d:%d\n",start.hours, start.minutes, start.seconds);
    printf("%d:%d:%d\n",t.hours, t.minutes, t.seconds);//因为 t 是全局变量，直接用 t 输出也可以

    p_start = split_time1(246800); // 方式二
    printf("%d:%d:%d\n",p_start->hours, p_start->minutes, p_start->seconds);
    
    return 0;
}