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法律研究C 程序将学生记录存储为结构，并按姓名对其进行排序
本题中，我们给出了一条包含学生 ID、学生姓名和学生成绩的学生记录。我们的任务是创建一个 C 程序，将学生记录存储为结构并按姓名对它们进行排序。
我们举个例子来理解一下这个问题，
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}}要解决此问题，我们首先将创建一个结构来存储学生详细信息。现在，我们将使用 qsort()，在 qsort 中我们将为 qsort 定义一个比较函数，该函数将使用 strcmp() 方法比较结构的名称。
实例
将学生记录存储为结构并按姓名对其进行排序的程序
//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];
//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
更新于： 2020 年 8 月 5 日
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