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