C 中的堆栈抽象数据类型 [英] Stack abstract data type in C

问题描述

你能告诉我我做错了什么吗?我收到 SIGSEGV(分段错误)错误.单链表是实现堆栈抽象数据类型的最佳方式吗?我尽量不使用全局变量，所以这就是我使用双指针的原因.

Can you please tell me what I did wrong? I'm getting SIGSEGV (Segmentation fault) error. Is single linked list the best way to implement a stack abstract data type? I'm trying not to use global variables so that's why I used double pointers.

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

typedef struct stack{
    int data;
    struct stack *next;
}STACK;

void push(STACK **head,STACK **tail,int n)
{
    STACK *x;
    if(*head==NULL)
    {
        (*head)=malloc(sizeof(STACK));
        (*head)->data=n;
        (*head)->next=NULL;
        *tail=*head;
    }
    else
    {
        x=malloc(sizeof(STACK));
        x->data=n;
        x->next=NULL;
        (*head)->next=x;
        (*head)=(*head)->next;
    }
}

void show(STACK *tail)
{
    if(tail!=NULL)
    {
        printf("From tail to head:\n");
        while(tail!=NULL)
        {
            printf("%d\n",tail->data);
            tail=tail->next;
        }
    }
    else
    {
        printf("The stack is empty!\n");
    }
}

void pop(STACK **head,STACK *tail)
{
    STACK *x;
    if(*head!=tail)
    {
        x=*head;
        while(tail->next->next!=NULL)
            tail=tail->next;
        printf("pop: %d\n",(*head)->data);
        *head=tail;
        free(x);
    }
    else
    {
        printf("pop: %d\n",(*head)->data);
        free(*head);
        *head=NULL;
    }
}

int main()
{
    STACK *head = NULL;
    STACK *tail = NULL;
    push(&head,&tail,4);
    pop(&head,tail);
    push(&head,&tail,7);
    push(&head,&tail,9);
    show(tail);
    return 0;
}

我编辑了代码，现在可以使用了.谢谢大家！！！

I edited the code, now it works. Thank you everyone!!!

推荐答案

最直接的问题是你永远不会在 main() 中初始化 head 和 tail:

The most immediate problem is that you never initialize head and tail in main():

STACK *head = NULL;
STACK *tail = NULL;

还有其他几个问题:

1. pop() 泄漏内存.
如果列表为空(即 tail 为 NULL)，
2. show() 将不起作用.
3. 当列表不为空时，show() 无法打印其元素之一.

1. pop() leaks memory.
2. show() won't work if the list is empty (i.e. tail is NULL).
3. When the list is not empty, show() fails to print one of its elements.

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