c - 使用指针在 C 中对结构进行排序

Question

我刚开始使用 C，对幕后发生的事情一无所知。我正在为一个数据结构类即时学习它，这让事情变得有点困难。

更新:我已经剥离了程序，从内存开始，然后再往上。我在那里有分配和解除分配函数，但出现 malloc 错误:Q1(9882) malloc: * error for object 0x7fff59daec08: pointer being freed was not allocated*在malloc_error_break中设置断点调试

Update2 这是我修改后的代码，它仍然缺少一些东西，我的几个 printf 语句没有显示:

#include <stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>

static int size = 10;

struct student{
    int id;
    int score;
};

struct student* allocate(){
     /*Allocate memory for ten students*/
     struct student *s = malloc(size*(sizeof(struct student)));
     assert(s != 0);
     /*return the pointer*/
     return s;
}

void generate(struct student* students){
    /*Generate random ID and scores for ten students, ID being between 1 and 10, scores between 0 and 100*/
    srand((unsigned int)time(NULL));
    int id[size];
    int y;

    for (int i = 0; i < size; i++){
        y = rand() % size + 1;
        while(dupe(id, i, y)){
            y = rand() % size + 1;
        }
        id[i] = y;
    }

    for (int j = 0; j < size; j++){
        (students + j)->id = id[j];
        (students + j)->score = rand() % 101;
        printf("ID: %d\tScore: %d\n", (students + j)->id, (students + j)->score);
    }
}

int dupe(int id[], int size1, int i){
    for (int x = 0; x < size1; x++){
        if(id[x] == i)
            return 1;
    }
    return 0;
}

void output(struct student* students){
     /*Output information about the ten students in the format:
              ID1 Score1
              ID2 score2
              ID3 score3
              ...
              ID10 score10*/
    sort(&students);
    for(int x = 0; x < size; x++){
        printf("ID: %d\tScore: %d\n", (students + x)->id, (students + x)->score); //print stmt not showing
    }
}

void sort(struct student* students){
    struct student *sd = allocate();

    struct student *stud;

    for(int i = 0; i < size; i++){
        stud = &students[i];
        sd[stud->id] = *stud;
    }
    for(int x = 0; x < size; x++){
        printf("ID: %d\tScore: %d\n", (sd + x)->id, (sd + x)->score); //print stmt not showing
    }
    students = &sd;
    deallocate(sd);
}

void summary(struct student* students){
     /*Compute and print the minimum, maximum and average scores of the ten students*/

}

void deallocate(struct student* stud){
     /*Deallocate memory from stud*/
    free(stud);
}

int main(){
    struct student* stud = NULL;
    char c[] = "------------------------------\n";
    /*call allocate*/
    stud = allocate();
    /*call generate*/
    generate(&stud);
    /*call output*/
    printf("%s", c);
    output(&stud);
    /*call summary*/

    /*call deallocate*/
    deallocate(stud);

    return 0;
}

Answer 1

students = &students[x];

这会改变 students 指向的位置，因此下一次循环时您将从那里开始偏移，而不是从头开始。也就是说，您将获得 originalstudents[0]、originalstudents[1]、originalstudents[1+2]、originalstudents[ 1+2+3] 等。sd 也有同样的问题。

相反，你想使用不同的变量，比如

struct student* st = &students[x];
printf("id = %d\tscore = %d\n", st->id, st->score);
etc

另外，sd 是做什么用的？您似乎无缘无故地分配了一些空间并将学生复制到 sd。分配的空间未保存或返回......这是内存泄漏。哦，等等，我明白了……你按照学生的 ID 对 sd 中的学生重新排序。所以你应该在完成后释放内存。但是对于学生和 sd，您需要一个指向数组的指针而不是指向数组元素的指针。您可以使用许多不同的命名约定，但最好使用一致的命名约定。例如:

void output(struct Student* students){
    struct Student *idstudents = allocate(); /* sorted by id */
    if (!idstudents)
        /* handle allocation error */;

    for (int x = 0; x < 10; x++){
        struct Student* student = &students[x];
        printf("id = %d\tscore = %d\n", student->id, student->score);
        struct Student* idstudent = &idstudents[student->id];
        *idstudent = *student; /* copy all fields at once */
        printf("id = %d\tscore = %d\n", idstudent->id, idstudent->score);/* pointless here, since we just printed the same info via student */
    }

    for (int x = 0; x < 10; x++){
        struct Student* idstudent = &idstudents[x];
        printf("id = %d\tscore = %d\n", idstudent->id, idstudent->score);
    }
    deallocate(idstudents);
}