Question

我这边有问题

typedef struct
{
  int hours;
  int minutes;
  int seconds; 
} Time;

Time calcTime(Time time1, Time time2)
{
    Time calc1;
    Time calc2;

    calc1.hours - calc2.hours;

    return calc1; // How do I print this outside the function?
}

如何打印 calc1.seconds OUTSIDE函数？

Answer 1

这就是你可以做到的。

Time t = calcTime(time1, time2);
printf("%d %d %d\n", t.hours, t.minutes, t.seconds);

但是，请使用time_t中的struct tm，time.h进行相关实施。您可以找到更多信息here或示例here。

calcTime

需要一些修正

Time calcTime(Time calc1, Time calc1)
{
    // Do not re declare the function arguments.
    //Time calc1;
    //Time calc2;

    // This has no effect, unless you store the result
    // calc1.hours - calc2.hours;
    calc1.hours -= calc2.hours;   // short for calc1.hours = calc1.hours - calc2.hours

    // you need to also check if hours has a valid value after the operation
    if(calc1.hours < 0 || calc1.hours > 24) {
        // error
    }

    return calc1;
}

Answer 2

即使您调用该函数，它也无法正常工作，因为您正在减去未初始化的变量。您在time1中不需要time2和calcTime，而只需像这样定义您的函数：

Time calcTime(Time calc1, Time calc2)
{
    //Time calc1;
    //Time calc2;

    // calc1.hours - calc2.hours; --> You are not storing the value here    

    calc1.hours -= calc2.hours; // or you should write --> calc1.hours = calc1.hours - calc2.hours

    return calc1; 
}

并致电main：

int main()
{
    Time t1,t2;
    t1.hours = 10;
    t2.hours = 5;
    t2 = calcTime(t1, t2);
    printf("The time is %d hours", t2.hours);
}

Answer 3

只需调用该函数（另请注意，我已将您的函数更改为实际返回相关值）。

typedef struct
{
  int hours;
  int minutes;
  int seconds; 
} Time;

Time calcTime(Time calc1, Time calc2)
{
    calc1.hours = calc1.hours - calc2.hours;

    return calc1;
}

int main()
{
    Time t1;
    Time t2;
    t1.hours = 10;
    t2.hours = 5;
    Time t3 = calcTime(t1, t2);
    printf("The time is %d hours", t3.hours);
}

从函数返回结构？

3 个答案: