为什么'->'运算符在这种指向指针的情况下不起作用？

Question

我尝试使用链表实现队列数据结构。 在函数 Enqueue 中，我需要返回前后指针的值，因为函数不能返回两个值，所以我使用了指向指针的指针：前指针和后指针。 运算符“->”似乎不起作用，我收到以下错误代码：

main.c:67:18: error: ‘*rear’ is a pointer; did you mean to use ‘->’?
             *rear->next=temp;
                  ^~
                  ->
main.c: In function ‘dequeue’:
main.c:78:14: error: ‘*front’ is a pointer; did you mean to use ‘->’?
      x=*front->data;
              ^~
              ->
main.c:80:19: error: ‘*front’ is a pointer; did you mean to use ‘->’?
      *front=*front->next;//correctly moving front pointer
                   ^~
                   ->

任何帮助将不胜感激。

//implementing queue
#include<stdio.h>
#include<stdlib.h>
typedef struct node {
    int data;
    struct node* next;
} *node;

//declaring functions
void createqueue();
void enqueue(node *front,node * rear, int x);
void dequeue(node *front, node * rear);
void display(node front);
void main(){
    node f=NULL;//pointer to a node
    node r=NULL;//pointer to a node
    
    char cont;
    cont='y';
    do{
        int x;
        int ch;
        printf("\n\nMain Menu:\n\n1.Create a Queue\n2.Enqueue an element onto the queue\n"
        "3.Dequeue an element of the queue\n4.Display the queue\n5.Quit\n");
        printf("\nEnter your choice: ");
        scanf("%d",&ch);
        switch(ch){
            case(1):
            createqueue();
            break;
            
            case(2):
            printf("Enter element : ");
            scanf("%d",&x);
            enqueue(&f,&r,x);
            break;
            
            case(3)://
            dequeue(&f, &r);
            break;
            
            case(4):
            display(f);
            break;
            
            case(5)://Quit
            exit(0);
        }
        printf("\nEnter 'y' or 'Y' to continue : ");
        scanf(" %c",&cont);
    }while(cont=='y'||cont=='Y');
}
void createqueue(){
    printf("Queue created!\n");
}
//the return values cannot be 2 in no. so we will have a pointer to the rear and front variables which are themselves pointers
void enqueue(node *front,node *rear,int x){//and not Enqueue(Node * front,Node * rear, int x){
    node temp=malloc(sizeof(struct node));
    if(temp!=NULL){
        temp->data=x;
        temp->next=NULL;
        //if both front and rear are null then both will change//temp and rear=null case:
        if (*front==NULL && *rear==NULL)
        *front=*rear=temp;//if there was only one element, the above case will happen for adding that first element and from the second element on, the else conditional would hold true
        //not empty queue case
        else{
            *rear->next=temp;
            *rear=temp;
        }
    }
    else
    printf("Unable to allocate memory for the node\n");
}
void dequeue(node *front,node *rear){
    int x;
    node t;
    if(*front!=NULL){//list not empty
        x=*front->data;
        t=*front;//t stores the address of the first node for freeing up later
        *front=*front->next;//correctly moving front pointer
        free(t);//free(address) is the syntax. free(**front) is not same as free(address of node) rather it is free(node) itself.
        if(*front==NULL)//if now after removing one element the list gets empty, then
        *rear=NULL;//otherwise rear will still be pointing to the address just freed
        printf("%d dequeued\n",x);
    }
    else
    printf("The list is empty!");
    
}
void display(node front){
    node t=front;
    int count=1;
    if(front!=NULL){
        while(t!=NULL){
            printf("%d. %d\n",count++,t->data);
            t=t->next;
        }
    }
    else
    printf("List is empty!");
}

Answer 1

也许你的函数声明应该更像

void 函数（节点*前，节点*后）；

有时修复语法中的小错误可以消除许多其他错误。

Answer 2

这是初学者犯的一个非常微妙的错误。 *rear->next=temp 用法错误 （*后）->下一个=温度；是正确的用法。 其他任务也类似。 这遵循运算符的优先级。

Answer 3

这是运算符的优先级问题：

*rear->next=temp 表示*(rear->next)=temp

但你需要的是：

(*rear)->next=temp

所以你必须明确地使用括号。

Answer 4

自从我使用 c++ 以来已经过去了二十年，但如果我看到你的错误我会尝试

*front = front->next

因为看起来您想将传入指针的值更改为列表中的下一项。