在其他结构中使用指针进行结构化分段错误

如何解决在其他结构中使用指针进行结构化分段错误

struct Tasks{

    int tid;                      
    int difficulty;               
    struct Tasks *next;           
};

struct Head_GL{

    int tasks_count[3];           
    struct Tasks *head;           
};

struct Head_GL *tasks_head;

，我必须创建一个具有添加顺序困难的链接列表。如何进行比较并阅读难度。我做了这个tasks_head->head->difficulty并给我分割错误

解决方法

您需要创建每个项目：

tasks_head = calloc(1,sizeof(struct Head_GL));
tasks_head->head = calloc(1,sizeof(struct Tasks));

然后可以填充并使用它们。您还需要记住以后释放它们。

我分配了内存，但是在打印结果时遇到问题。 main（tasks_head-> head-> tid）中的printf中的段错误。有帮助吗？

struct Tasks *new=(struct Tasks*)malloc(sizeof(struct Tasks));


tasks_head=(struct Head_GL*)malloc(sizeof(struct Head_GL));
tasks_head->head=(struct Tasks*)malloc(sizeof(struct Tasks));

tasks_head->tasks_count[0]=0;
tasks_head->tasks_count[1]=0;
tasks_head->tasks_count[2]=0;
tasks_head->head->difficulty=0;
tasks_head->head->tid=0;
tasks_head->head->next=NULL;

if(new==NULL)
    return 0;
new->tid = tid;
new->difficulty = difficulty;
new->next = NULL;

if(difficulty==1)
    tasks_head->tasks_count[0]++;
else if(difficulty==2)
    tasks_head->tasks_count[1]++;
else
    tasks_head->tasks_count[2]++;


if(tasks_head==NULL){
    tasks_head->head = new;
    return 1;
}
if( tasks_head->head->difficulty > difficulty){
    new->next = tasks_head->head;
    tasks_head->head= new;
    return 1;
}
else{
    prev = tasks_head->head;
    temp = tasks_head->head->next;
    while(temp != NULL && temp->difficulty < difficulty){
        prev = temp;
        temp = temp->next;
    }
    if(temp==NULL){
        prev->next = new;
        return 1;
    }
    else{
        new->next = temp;
        prev->next = new;
        return 1;
    }
}

}

int main（）{

printf("hello1\n");

if(1==insert_task(1,1))
    printf("alo");

if(1==insert_task(4,1))
    printf("alo");

if(1==insert_task(3,2))
    printf("alo\n");


printf("%d\n",num);

for(int i=0; i<num; i++){
    printf("%d\n",tasks_head->head->tid);
    tasks_head->head=tasks_head->head->next;
}

return 0;

}