我正在编写一个程序来计算两个给定时间之间耗时。
出于某种原因,我收到错误:在我的主要函数之前,我的 elapsedTime 函数原型(prototype)需要标识符或“C”。
我试过在程序中移动它,如果我在声明 t1 和 t2 之后找到它也没有什么不同。有什么问题?
谢谢
#include <stdio.h>
struct time
{
int seconds;
int minutes;
int hours;
};
struct elapsedTime(struct time t1, struct time t2);
int main(void)
{
struct time t1, t2;
printf("Enter start time: \n");
printf("Enter hours, minutes and seconds respectively: ");
scanf("%d:%d:%d", &t1.hours, &t1.minutes, &t1.seconds);
printf("Enter stop time: \n");
printf("Enter hours, minutes and seconds respectively: ");
scanf("%d:%d:%d", &t2.hours, &t2.minutes, &t2.seconds);
elapsedTime(t1, t2);
printf("\nTIME DIFFERENCE: %d:%d:%d -> ", t1.hours, t1.minutes, t1.seconds);
printf("%d:%d:%d ", t2.hours, t2.minutes, t2.seconds);
printf("= %d:%d:%d\n", differ.hours, differ.minutes, differ.seconds);
return 0;
}
struct elapsedTime(struct time t1, struct time t2)
{
struct time differ;
if(t2.seconds > t1.seconds)
{
--t1.minutes;
t1.seconds += 60;
}
differ.seconds = t2.seconds - t1.seconds;
if(t2.minutes > t1.minutes)
{
--t1.hours;
t1.minutes += 60;
}
differ.minutes = t2.minutes - t1.minutes;
differ.hours = t2.hours - t1.hours;
return differ;
}
最佳答案
您的函数没有正确定义返回类型:
struct elapsedTime(struct time t1, struct time t2);
struct
本身不足以定义返回类型。您还需要结构名称:
struct time elapsedTime(struct time t1, struct time t2);
您还需要将函数的返回值分配给某些东西:
struct time differ = elapsedTime(t1, t2);
有了这个工作,你在做差异时“借用”的逻辑是倒退的:
if(t1.seconds > t2.seconds) // switched condition
{
--t2.minutes; // modify t2 instead of t1
t2.seconds += 60;
}
differ.seconds = t2.seconds - t1.seconds;
if(t1.minutes > t2.minutes) // switched condition
{
--t2.hours; // modify t2 instead of t1
t2.minutes += 60;
}
照原样,如果 t1
在 t2
之后,小时将为负数。如果您认为这意味着结束时间是第二天,则将小时数增加 24:
if(t1.hours > t2.hours)
{
t2.hours+= 24;
}
differ.hours= t2.hours - t1.hours;
关于c - 我调用这个函数错了吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38511912/