如何将结构转换为工作数组 C

Question

I have little programming experience and have had some issues regarding trying to get said structure to be read as an array.

The code that I have been working on does not seem to output anything at the moment. I want to eventually be able to sort the array, but first I need to actually get an array going.

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>

 struct Students {
    char name[20];
    char surname[20];
    int id;
    float lab;
    float kd;
 };

 void printStudent(struct Students *st );
 float student(struct Students *st);

 int main() {

 float student(struct Students *st){

    struct Students stud[4];

    strcpy(stud[0].name, "John");
    strcpy(stud[0].surname, "Doe");
    stud[0].id = 456;
    stud[0].lab = 7.31;
    stud[0].kd = 8.55;

    strcpy(stud[1].name, "Tom");
    strcpy(stud[1].surname, "K");
    stud[1].id = 001;
    stud[1].lab = 9.25;
    stud[1].kd = 8.93;

    strcpy(stud[2].name, "Jane");
    strcpy(stud[2].surname, "Dee");
    stud[2].id = 201;
    stud[2].lab = 7.31;
    stud[2].kd = 6.55;

    strcpy(stud[3].name, "Alice");
    strcpy(stud[3].surname, "Lee");
    stud[3].id = 199;
    stud[3].lab = 8.95;
    stud[3].kd = 9.22;

    strcpy(stud[4].name, "Victor");
    strcpy(stud[4].surname, "Mann");
    stud[4].id = 167;
    stud[4].lab = 10.00;
    stud[4].kd = 9.50;

    printStudent(&st[0]);
    printStudent(&st[1]);
    printStudent(&st[2]);
    printStudent(&st[3]);
    printStudent(&st[4]);


    return 0;
 }

 void printStudent(struct Students *st){

    printf("%s \n", st->name);
    printf("%s \n", st->surname);
    printf("%d \n", st->id);
    printf("%.2f \n", st->lab);
    printf("%.2f \n", st->kd);
 }
 }

Answer 1

You want this:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Students {
  char name[20];
  char surname[20];
  int id;
  float lab;
  float kd;
};

void printStudent(struct Students* st);
float student(struct Students* st);

void printStudent(struct Students* st) {

    printf("%s \n", st->name);
    printf("%s \n", st->surname);
    printf("%d \n", st->id);
    printf("%.2f \n", st->lab);
    printf("%.2f \n", st->kd);
}


  int main() {
    struct Students stud[5];    // you need 5 students (0, 1, 2, 3, 4) that's 5

    // fill students 0 to 4

    strcpy(stud[0].name, "John");
    strcpy(stud[0].surname, "Doe");
    stud[0].id = 456;
    stud[0].lab = 7.31;
    stud[0].kd = 8.55;

    strcpy(stud[1].name, "Tom");
    strcpy(stud[1].surname, "K");
    stud[1].id = 001;
    stud[1].lab = 9.25;
    stud[1].kd = 8.93;

    strcpy(stud[2].name, "Jane");
    strcpy(stud[2].surname, "Dee");
    stud[2].id = 201;
    stud[2].lab = 7.31;
    stud[2].kd = 6.55;

    strcpy(stud[3].name, "Alice");
    strcpy(stud[3].surname, "Lee");
    stud[3].id = 199;
    stud[3].lab = 8.95;
    stud[3].kd = 9.22;

    strcpy(stud[4].name, "Victor");
    strcpy(stud[4].surname, "Mann");
    stud[4].id = 167;
    stud[4].lab = 10.00;
    stud[4].kd = 9.50;

    // print students 0 to 4
    printStudent(&stud[0]);
    printStudent(&stud[1]);
    printStudent(&stud[2]);
    printStudent(&stud[3]);
    printStudent(&stud[4]);

    return 0;
  }

There is not much to comment, it's pretty autodocumented.