功能比较两个约会 [英] Function to compare two appointment

问题描述

用户将输入两个约会



比较的计算是否冲突

User will input two appointments

What is the calculation to compare is conflict or not

int main(){

    struct Appointment aptOne, aptTwo;
    make_appointment(&aptOne, "Appointment One",
    				2018, 1, 23, 11, 30, 12, 45);
    
    make_appointment(&aptTwo, "Appointment Two",
    				2018, 1, 23, 12, 45, 13, 15);


}

我的尝试：

What I have tried:

引用：

试图拥有一个函数和布尔值

trying to have a function and boolean

struct Appointment {
    char description;
    int year;
    int month;
    int day;
    int startHour;
    int startMinute;
    int endHour;
    int endMinute;
};

bool has_conflict( struct Appointment* pAptA,
                  struct Appointment* pAptB)
{
   return true;

}


}

int main(){

    -----

    if (has_conflict( &aptOne, &aptTwo)){
    		printf("Appointments CLASH!\n");
    } else {
    		printf("Appointments Fine!\n");
    }
}

推荐答案

想一想：你会在真实的中做些什么世界？

你会说：约会A从X开始，到Y结束。预约B的开始是否属于那个？预约B的结束是否属于那个？

如果任何一个问题都是真的，它们会重叠。



但是...制造东西整个负载更容易，不要存储这样的约会。

有一个名为 time_t [ ^ ]哪些商店一个日期和时间值，其中包括比较功能，这将使您的生活更轻松 - 特别是如果约会可能跨越午夜。

如果您在约会中存储两个time_t值来表示开始和约会的终点，您可以使用 difftime [ ^ ]非常容易比较两个完整的日期时间对象。

Think about it: what would you do in the "real world"?
You'd say: Appointment A starts at X, and end at Y. Does the start of Appointment B fall inside that? Does the end of Appointment B fall inside that?
If either question is true, they overlap.

But... to make things a whole load easier, don't store your appointments like that.
There is a struct called time_t[^] which stores a date and time value, and which includes comparison functions which would make your life a whole load easier - particularly if an appointment might span midnight.
If you store two time_t values in your Appointment to represent the start and end point of the appointment, you can use difftime[^] to compare the two whole datetime objects very easily.

以愚蠢的方式做到这一点mpare每个元素

do it in a dumb way compare each of the element

bool has_conflict( struct Appointment* pAptA,
                  struct Appointment* pAptB)
{
    if((*pAptA).year == (*pAptB).year &&
       (*pAptA).month == (*pAptB).month &&
       (*pAptA).day == (*pAptB).day &&
       (*pAptA).startHour == (*pAptB).startHour &&
       (*pAptA).startMinute == (*pAptB).startMinute &&
       (*pAptA).endHour == (*pAptB).endHour &&
       (*pAptA).endMinute == (*pAptB).endMinute){
        return true;
    }else {return false;}

这篇关于功能比较两个约会的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！