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法律研究C 程式將學生記錄儲存為結構，並按姓名對其進行排序
本題中，我們給出了一條包含學生 ID、學生姓名和學生成績的學生記錄。我們的任務是建立一個 C 程式，將學生記錄儲存為結構並按姓名對它們進行排序。
我們舉個例子來理解一下這個問題，
輸入 - 學生記錄 =
{{ 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}}輸出 - 學生記錄 =
{{ 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}}要解決此問題，我們首先將建立一個結構來儲存學生詳細資訊。現在，我們將使用 qsort()，在 qsort 中我們將為 qsort 定義一個比較函式，該函式將使用 strcmp() 方法比較結構的名稱。
例項
將學生記錄儲存為結構並按姓名對其進行排序的程式
//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;
}輸出
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
更新於： 2020 年 8 月 5 日
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