在这个问题中,我们得到了一个包含student_id,student_name,student_percentage的学生记录。我们的任务是创建一个C程序,以结构形式存储学生记录并按名称对其进行排序。
让我们举个例子来了解这个问题,
输入-学生记录=
{{ student_id = 1, student_name = nupur, student_percentage = 98},
{ student_id = 2, student_name = Akash, student_percentage = 75},
{ student_id = 3, student_name = Yash, student_percentage = 62},
{ student_id = 4, student_name = Jyoti, student_percentage = 87},
{ student_id = 5, student_name = Ramlal, student_percentage = 80}}输出-学生记录=
{{ student_id = 2, student_name = Akash, student_percentage = 75},
{ student_id = 4, student_name = Jyoti, student_percentage = 87},
{ student_id = 1, student_name = nupur, student_percentage = 98},
{ student_id = 5, student_name = Ramlal, student_percentage = 80},
{ student_id = 3, student_name = Yash, student_percentage = 62}}为了解决这个问题,我们将首先创建一个结构,该结构将存储学生的详细信息。现在,我们将使用,qsort()并且在该qsort中,我们将为此qsort定义一个比较器函数,该函数将使用strcmp()方法比较结构的名称。
该程序将学生记录存储为结构并按名称对其进行排序
//C program to store Student records as Structures and Sort them by Name
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Student {
int student_id;
char* student_name;
int student_percentage;
};
int comparator(const void* s1, const void* s2){
return strcmp(((struct Student*)s1)->student_name,((struct Student*)s2)->student_name);
}
int main() {
int n = 5;
struct Student arr[n];
//学生1-
arr[0].student_id = 1;
arr[0].student_name = "Nupur";
arr[0].student_percentage = 98;
//学生2-
arr[1].student_id = 2;
arr[1].student_name = "Akash";
arr[1].student_percentage = 75;
//学生3-
arr[2].student_id = 3;
arr[2].student_name = "Yash";
arr[2].student_percentage = 62;
//学生4-
arr[3].student_id = 4;
arr[3].student_name = "Jyoti";
arr[3].student_percentage = 87;
//学生5-
arr[4].student_id = 5;
arr[4].student_name = "Ramlal";
arr[4].student_percentage = 80;
printf("Unsorted Student Record:\n");
for (int i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n", arr[i].student_id, arr[i].student_name, arr[i].student_percentage);
}
qsort(arr, n, sizeof(struct Student), comparator);
printf("\n\nStudent Records sorted by Name:\n");
for (int i = 0; i < n; i++) {
printf("Id = %d, Name = %s, Age = %d \n", arr[i].student_id, arr[i].student_name, arr[i].student_percentage);
}
return 0;
}输出结果
Unsorted Student Record: Id = 1, Name = Nupur, Age = 98 Id = 2, Name = Akash, Age = 75 Id = 3, Name = Yash, Age = 62 Id = 4, Name = Jyoti, Age = 87 Id = 5, Name = Ramlal, Age = 80 Student Records sorted by Name: Id = 2, Name = Akash, Age = 75 Id = 4, Name = Jyoti, Age = 87 Id = 1, Name = Nupur, Age = 98 Id = 5, Name = Ramlal, Age = 80 Id = 3, Name = Yash, Age = 62