c - 'struct queue' 没有名为 'next' 的成员。

Question

我尝试使用以下结构使用链表实现队列。但是在入队函数中，就在评论下方，我收到了上述错误。此错误已发生，同时在此链接给出的代码中使用了相同的代码 http://quiz.geeksforgeeks.org/queue-set-2-linked-list-implementation/ .有一条评论是我用来运行代码的。但我不明白为什么我会收到此错误。请帮忙!

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

    struct node
    {
        int data;
        struct node *next;
    };

    struct queue
    {
        struct queue *front;
        struct queue *rear;
    };

    struct queue* createQueue();
    void enqueue(struct queue *q,int info);
    int dequeue(struct queue *q);
    void display(struct queue *q);
    int isEmpty(struct queue *q);
    void menu();

    struct queue* createQueue()
    {
        struct queue *new_queue; 
        new_queue = (struct queue*) malloc(sizeof(struct queue));
        if(new_queue != NULL)
        {
            return ;
        } 
        new_queue->front = new_queue->rear = NULL;
        return new_queue;
    }
    int isEmpty(struct queue *q)
    {
        if(q->front == NULL)
            return 1;
    }

    void enqueue(struct queue *q,int info)
    {
        struct node *temp,*temp1;
        temp = (struct node*) malloc(sizeof(struct node));
        if(temp == NULL)
            return ;
        temp->data = info;
        temp->next = NULL;
        if(q->front == NULL && q->rear == NULL)
        {
            q->front = temp;
            q->rear = temp;
        }
        if(q->rear != NULL)
        {
            //temp1 = q->rear;
            q->rear->next = temp;
            q->rear = temp;
        }
    }

    int dequeue(struct queue *q)
    {
        if(isEmpty(q) == 1)
        {
            printf("Empty list!");
        }
        struct node *temp = q->front;
        q->front = temp->next;
        int data = temp->data;
        free(temp);
        return (data);
    }

    void display(struct queue *q)
    {
        struct node *start,*end = NULL;
        start = q->front;
        end = q->rear;
        if(isEmpty(q) == 1)
        {
            printf("Empty List!");
        }
        while(start != NULL)
        {
            printf("%d",start->data);
            start = start->next;
        }
    }

    void menu()
    {
        int choice = 0;
        int input = 0;
        struct queue *q;
        while(1)
        {
            printf("\t\tMain Menu\n");
            printf("\t0.Create Queue\n");
            printf("\t1.Enqueue\n");
            printf("\t2.Dequeue\n");
            printf("\t3.Display\n");
            printf("\t4.Exit\n");
            printf("\tEnter the desired choice:");
            scanf("%d",&choice);
            switch(choice)
            {
                case 0:
                    q = createQueue();
                    break;
                case 1:
                    printf("\tEnter the data:");
                    scanf("%d",&input);
                    enqueue(q,input);
                    break;
                case 2:
                    printf("%d",dequeue(q));
                    break;
                case 3:
                    display(q);
                    break;
                case 4: 
                    exit(0);
                default:
                    printf("\tThe value is invalid!");
            }
        }

    }

    int main()
    {
        menu();
        return 0;
    }

Answer 1

你在写

q->rear->next = temp;

rear 是一个 queue 结构，而不是一个 node 结构。

你写了这个:

struct queue
{
    struct queue *front;
    struct queue *rear;
};

但从你使用的术语来看，我认为它应该是这样的:

struct queue
{
    struct node *front;
    struct node *rear;
};

(当然错误行会变得正常，但你可能还有其他问题:))