c - 用指针对C中的结构进行排序

我刚从C开始，对幕后的情况一无所知。我正在为一个数据结构类动态学习，这使事情变得更难。
更新：我已经将程序剥离下来，从内存和启动。我在那里有allocate和deallocate函数，得到了一个malloc错误：Q1（9882）malloc:*对象0x7fff59daec08的错误：未分配被释放的指针
*在malloc\u error\u break中设置要调试的断点
Update2这是我修改过的代码，它仍然缺少一些内容，我的printf语句没有出现：

#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;
}

最佳答案

students = &students[x];

这会改变students点的位置，因此下次通过循环时，您将从那里偏移，而不是从一开始偏移。也就是说，你得到了originalstudents[0]，originalstudents[1]，originalstudents[1+2]，originalstudents[1+2+3]等等。你也有同样的问题。
相反，您需要使用一个不同的变量，比如

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

还有，sd是干什么的？你似乎分配了一些空间，并没有明显的原因将学生复制到sd。分配的空间没有保存或返回。。。这是内存泄漏。哦，等等，我明白了。。你按照学生的身份重新排列学生的顺序，这样你就可以在完成后释放记忆。但是对于student和sd，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);
}