{{ 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，我们首先会创建一个存储学生详细信息的结构。现在，我们将使用qsort()函数，并在该函数中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。

示例

将学生记录存储为结构并按名称排序的程序

在线演示

//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];
   //student 1
   arr[0].student_id = 1;
   arr[0].student_name = "Nupur";
   arr[0].student_percentage = 98;
   //student 2
   arr[1].student_id = 2;
   arr[1].student_name = "Akash";
   arr[1].student_percentage = 75;
   //student 3
   arr[2].student_id = 3;
   arr[2].student_name = "Yash";
   arr[2].student_percentage = 62;
   //student 4
   arr[3].student_id = 4;
   arr[3].student_name = "Jyoti";
   arr[3].student_percentage = 87;
   //student 5
   arr[4].student_id = 5;
   arr[4].student_name = "Ramlal";
   arr[4].student_percentage = 80;
   printf("Unsorted Student Record:
");
   for (int i = 0; i < n; i++) {
      printf("Id = %d, Name = %s, Age = %d 
", arr[i].student_id, arr[i].student_name, arr[i].student_percentage);
   }
   qsort(arr, n, sizeof(struct Student), comparator);
   printf("
Student Records sorted by Name:
");
   for (int i = 0; i < n; i++) {
      printf("Id = %d, Name = %s, Age = %d 
", arr[i].student_id, arr[i].student_name, arr[i].student_percentage);
   }
   return 0;
}

输出

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