1

我有一个嵌套结构，“内部结构”是一个带有成员的日期结构；日、月、年。这些结构包含在一个动态数组中。我想遍历结构并找出哪个结构具有最旧的日期。我是编程新手，不太确定如何处理这个问题。请帮忙。谢谢！

#include <stdio.h>
#include <stdlib.h>

//define structure to store students birth date
struct date
{
    int month;
    int day;
    int year;
};
//define structure to store student info and date structure for birth date
struct studentInfo
{
    int iD;
    struct date birthDate;
    int phone;
};

int main(void)
{
    //declare and initialize variables
    int recNum = 0;     //number of records
    struct studentInfo * records = NULL;    //struct pointer, array
    //request user input  and store in recNum for record amount
    printf("\nHow many students do you need to enter records for?:");
    scanf ("%d",&recNum);
    //dynamically allocate memory
    records = (struct studentInfo*)malloc((sizeof(struct studentInfo)*recNum));
    //loop through records and request/store values from user
    int count;
    int studentNum=1;
    for(count=0;count<recNum;count++)
    {
        printf("Please enter the following for student number %d\n",studentNum);
        //request and store student ID
        printf("Student ID#:");
        scanf ("%d",&records[count].iD);
        //request and store student phone number
        printf("Student phone# (numbers only, 10 digits):");
        scanf ("%d",&records[count].phone);

        //error checking, check if phone number is 10 digits
        int phoneCount = 0;
        int phoneCopy = records[count].phone;
        while(phoneCopy != 0)
        {
            phoneCopy /= 10;
            phoneCount++;
        }
        if (phoneCount != 10)
        {
            printf("The number you have entered is not 10 digits, please re-enter:");
            scanf ("%d",&records[count].phone);
        }

        //request and store student birthdate
        printf("Student birthday (mm/dd/yyyy):");
        scanf("%d/%d/%d",&records[count].birthDate.month,&records[count].birthDate.day,
                &records[count].birthDate.year);

        //test stuff
        printf("Student number %d has an ID of %d and a phone number of %d\n", studentNum,
                records[count].iD, records[count].phone);
        studentNum++;
    }
    return 0;
}
4

2 回答 2

0 
#define NUMCMP(x,y) (((x) < (y)) ? -1 : ((x) > (y)) ? 1 : 0)

int compar_student_byDate(const struct studentInfo *student1, const struct studentInfo *student2){
    struct date date1 = student1->birthDate;
    struct date date2 = student2->birthDate;
    int tmp;
    if((tmp=NUMCMP(date1.year, date2.year))==0){
        if((tmp=NUMCMP(date1.month, date2.month))==0)
            return NUMCMP(date1.day, date2.day);
        else
            return tmp;
    } else
        return tmp;
}

struct studentInfo *oldest(struct studentInfo *records, int recNum){
    struct studentInfo *old = records;
    int i;
    for(i = 1;i<recNum;++i){
        if(compar_student_byDate(old, &records[i])>0)
            old = &records[i];
    }
    return old;
}

在主要

struct studentInfo *old_student = oldest(records, recNum);
于 2013-07-04T22:35:13.243 回答
-1

好的，首先我应该告诉你，这是一个可以有很多很多正确答案的问题。您可以创建一个树而不是创建一个结构数组，studentInfo并且每当您将一个新节点插入到树中时，将根节点保持为许多其他解决方案中最旧的。无论如何，对于您的简单案例，一个简单的解决方案可能如下

步骤 0. 你想找到

studentInfo *oldest = records;

步骤 1. 遍历数组

studentInfo *traveler;
for(traveler = records; i<recNum; i++, traveler++)

第 2 步。比较最老和旅行者的日期。

if(oldest->birthDate.year > traveler->birthDate.year && 
   oldest->birthDate.month > traveler->birthDate.month &&
   oldest->birthDate.day > traveler->birthDate.day)

步骤 3. 每次找到一个时更新您找到的最旧的记录

oldest = traveler;

完成迭代后，最旧的记录应该是您要查找的记录。

于 2013-07-04T21:58:14.127 回答