为什么不是我的 reverse();功能工作？

如何解决为什么不是我的 reverse();功能工作？

我正在用 C 编写一个程序，用于反转循环单向链表。由于某种原因，我不断收到分段错误。我确定问题出在 reverse 函数上，因为我尝试注释函数调用，程序运行正常。

对于我的 reverse() 函数，我使用了 3 个指针：prev、next 和 curr。逻辑是我将运行一个循环，直到 curr 获取 head 的地址，该地址将存储在最后一个节点的 link 部分，因为它是一个循环链表。我将使用 curr->link 指针不断更新 prev，这会将其链接从下一个节点更改为上一个节点。

当循环中断时，head->link = prev; 和 head = prev; 将更新各自的地址，使它们指向反向列表的第一个节点。

//reversing CLL

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

struct node {
    int data;
    struct node *link;
} *head;

void reverse() {
    struct node *prev = NULL,*curr = head,*next;
        
    while (curr != head) {
        next = curr->link;
        curr->link = prev;
        prev = curr;
        curr = next;
    }
        
    head->link = prev;
    head = prev; 
}

void createList(int n) {
    int i,data;    
    
    head = (struct node *)malloc(sizeof(struct node));
        
    struct node *ptr = head,*temp;
            
    printf("Enter data of node 1\t");
    scanf("%d",&data);
            
    head->data = data;
    head->link = NULL;
            
    for (i = 2; i <= n; i++) {
        temp = (struct node *)malloc(sizeof(struct node));
                        
        printf("Enter data of node %d\t",i);
        scanf("%d",&data);
                        
        temp->data = data;
        temp->link = NULL;
                        
        ptr->link = temp;
        ptr = ptr->link;
    }
    ptr->link = head;
}

void disp() {
    struct node *ptr = head;
        
    do {
        printf("%d\t",ptr->data);   //gdb debugger shows problem is in this line
        ptr = ptr->link;
    } while (ptr != head);
}

int main() {
    int n;
        
    printf("Enter no of nodes to be created\t");
    scanf("%d",&n);
        
    createList(n);
            
    printf("\n\nList is displayed below!\n");
        
    disp();
            
    printf("\n\nreversing list ...\n");
            
    reverse();   // on commenting this call,disp() function 
                 // works accurately showing node data non-reversed
                  
    disp();
            
    printf("\n\nList successfully reversed!\n");
}

解决方法

reverse() 函数中的循环立即退出，因为 curr 被初始化为 head 的值，因此测试 while (curr != head) 在第一次迭代时为假。

reverse() 然后将 head->link 设置为 NULL，最后 head 也设置为 NULL（prev 的初始值），即解释了后续 disp() 函数中的分段错误，其中您使用了无法处理空列表的 do { } while (pre != head)。

这是修改后的版本：

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

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

struct node *reverse(struct node *head) {
    struct node *prev = NULL,*curr = head;

    if (head) {
        do {
            struct node *next = curr->link;
            curr->link = prev;
            prev = curr;
            curr = next;
        } while (curr != head);
        curr->link = prev;
        head = prev;
    }
    return head;
}

struct node *createList(int n) {
    struct node *head = NULL,*tail = NULL,*temp;
    int i;

    for (i = 1; i <= n; i++) {
        temp = (struct node *)malloc(sizeof(struct node));
        temp->data = 0;
        temp->link = NULL;

        printf("Enter data of node %d\t",i);
        scanf("%d",&temp->data);

        if (head == NULL) {
            head = temp;
        } else {
            tail->link = temp;
        }
        tail = temp;
        temp->link = head;
    }
    return head;
}

void disp(struct node *head) {
    if (head) {
        struct node *ptr = head;
        do {
            printf("%d\t",ptr->data);
            ptr = ptr->link;
        } while (ptr != head);
    }
}

int main() {
    struct node *head;
    int n = 0;

    printf("Enter no of nodes to be created\t");
    scanf("%d",&n);

    head = createList(n);

    printf("\n\nList is displayed below!\n");
    disp(head);

    printf("\n\nReversing list ...\n");

    head = reverse(head);

    disp(head);

    printf("\n\nList successfully reversed!\n");

    // should free the list
    return 0;
}

,

对于初学者来说，使用全局变量 YYYY-MM-DD

head

在这种情况下，函数依赖于全局变量，并且您不能在一个程序中使用多个列表。

由于这个初始化

struct node {
    int data;
    struct node *link;
} *head;

while 循环的条件

struct node *prev = NULL,*curr = head,*next;
                          ^^^^^^^^^^^^

is 永远不会评估为真，因为最初指针 while (curr != head) { 等于指针 curr。

此外，如果列表为空，则此语句

head

调用未定义的行为。

这是一个演示程序，展示了如何在 main 中声明列表然后反转。

head->link = prev;

程序输出为

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

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

size_t assign( struct node **head,const int a[],size_t n )
{
    while ( *head )
    {
        struct node *tmp = *head;
        *head = ( *head )->link;
        free( tmp );
    }
    
    size_t total = 0;
    
    struct node *first = NULL;
    
    while ( total < n && ( *head = malloc( sizeof( struct node ) ) ) != NULL )
    {
        ( *head )->data  = a[total];
        ( *head )->link  = NULL;
        ++total;
        
        if ( first == NULL ) first = *head;
        
        head = &( *head )->link;
    }
    
    if ( first != NULL ) *head = first;
    
    return total;
}

void display( const struct node *head )
{
    if ( head != NULL )
    {
        const struct node *current = head;
        do
        {
            printf( "%d -> ",current->data );
        } while ( ( current = current->link ) != head );
    }       
    
    puts( "null" );
}

struct node * reverse( struct node **head )
{
    if ( *head )
    {
        struct node *last = *head;
        struct node *prev = NULL;

        while ( ( *head )->link != last )
        {
            struct node *current = *head;
            *head = ( *head )->link;
            current->link = prev;
            prev = current;
        }
        
        ( *head )->link = prev;
        last->link = *head;
    }
    
    return *head;
}

int main(void) 
{
    struct node *head = NULL;
    
    int a[] = { 0,1,2,3,4,5,6,7,8,9 };
    
    assign( &head,a,sizeof( a ) / sizeof( *a ) );
    
    display( head );
    
    display( reverse( &head ) );
    
    display( reverse( &head ) );
    
    return 0;
}