c - 如何解决以下代码中的逻辑错误

Question

我在执行以下代码时遇到问题

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


struct linked
{
    int val;
    struct linked *before;
    struct linked *next;
};

void get(struct linked *var);
void printforward(struct linked *var);
void printbackward(struct linked *var);



int main()
{
    struct linked *ptr,*temp;
    ptr=(struct linked*)malloc(sizeof(struct linked));
    if(ptr==NULL)
    {
        printf("NOT ENOUGH MEMOREY");
        exit(2);
    }
    ptr->before=NULL;
    get(ptr);
    printf("\nForward:\n");
    printforward(ptr);
    for(temp=ptr;temp->next;temp=temp->next)
    {
        temp->next->before=(struct linked*)temp;
    }
    printf("\nBackward:\n");
    printbackward(temp->before);
 }


void get(struct linked *var)
 {
    printf("Enter the number (-99) to quit:");
    scanf("%d",&var->val);
    if(var->val==-99)
    {
         var->next=NULL;
         return;
     }
   else
   {
        var->next=(struct linked*)malloc(sizeof(struct linked));
        if(var->next==NULL)
        {
            printf("NOT ENOUGH MEMOREY");
            exit(2);
         }
         get(var->next);
   }
}

void printforward(struct linked *var)
{
    if(var->next==NULL)
   {
       return;
   }
   else
    {
        printf("\n%d",var->val);
        printforward(var->next);
    }
}

 void printbackward(struct linked *var)
 {
     if(var->before==NULL)
     {
         printf("\n%d",var->val);
         return;
     }
     else
     {
         printf("\n%d",var->val);
         printforward(var->before);
     }
    }

输出:

 Enter the number (-99) to quit:1
 Enter the number (-99) to quit:2
 Enter the number (-99) to quit:3
 Enter the number (-99) to quit:4
 Enter the number (-99) to quit:5
 Enter the number (-99) to quit:6
 Enter the number (-99) to quit:7
 Enter the number (-99) to quit:8
 Enter the number (-99) to quit:9
 Enter the number (-99) to quit:0
 Enter the number (-99) to quit:-99

Forward:

1
2
3
4
5
6
7
8
9
0
Backward:

0
9
0
Process returned 0 (0x0)   execution time : 22.297 s
Press any key to continue.

预期输出:

Enter the number (-99) to quit:1
Enter the number (-99) to quit:2
Enter the number (-99) to quit:3
Enter the number (-99) to quit:4
Enter the number (-99) to quit:5
Enter the number (-99) to quit:6
Enter the number (-99) to quit:7
Enter the number (-99) to quit:8
Enter the number (-99) to quit:9
Enter the number (-99) to quit:0
Enter the number (-99) to quit:-99

Forward:

1 
2
3
4
5
6
7
8
9
0
Backward:
0
9
8
7
6
5
4
3
2
1

让我知道代码中有什么问题，我正在学习c语言的链表，我想编写双向链表的代码，但是我在上面的问题中有一个逻辑错误，我不明白为什么上面的代码不起作用，所以请求消除疑问。

Answer 1

为什么你要通过递归来做到这一点，我不知道，但本着这种精神，请考虑这一点。

• 您正在从 printbackward 调用 printforward，这是没有意义的。
• 你的 Actor 阵容没有帮助，事实上，他们掩盖了真正的问题。在 C 中，既不需要也不建议进行内存分配、like-pointer 或往返于 void-pointer 的转换。 Read here for why

所有四个操作都可以递归完成，事实上，您甚至不需要“before”指针，但我仍然保留了它。您可以:

• 建立您的 list
• 转发打印您的列表
• 向后打印您的列表
• 清理您的列表

...全部使用递归(我将您想要的原因作为不同的问题)。

<小时/>

代码

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

struct linked
{
    int val;
    struct linked *before;
    struct linked *next;
};

struct linked *get(struct linked *before);
void printforward(struct linked const *var);
void printbackward(struct linked const *var);
void cleanup(struct linked *lst);

int main()
{
    struct linked *ptr = get(NULL);

    printf("Forward: ");
    printforward(ptr);
    fputc('\n', stdout);

    printf("Backward: ");
    printbackward(ptr);
    fputc('\n', stdout);

    cleanup(ptr);
}


struct linked *get(struct linked *before)
{
    printf("Enter the number (-99) to quit:");
    int value = -99;
    if (scanf("%d", &value) != 1 || value == -99)
        return NULL;

    struct linked *p = malloc(sizeof *p);
    if (p != NULL)
    {
        p->before = before;
        p->val = value;
        p->next = get(p);
    }
    return p;
}

void printforward(struct linked const *var)
{
    if (var)
    {
        printf("%d ", var->val);
        printforward(var->next);
    }
}

void printbackward(struct linked const *var)
{
    if (var)
    {
        printbackward(var->next);
        printf("%d ", var->val);
    }
}

void cleanup(struct linked *lst)
{
    if (!lst)
        return;

    cleanup(lst->next);
    free(lst);
}

控制台

Enter the number (-99) to quit:1
Enter the number (-99) to quit:2
Enter the number (-99) to quit:3
Enter the number (-99) to quit:4
Enter the number (-99) to quit:5
Enter the number (-99) to quit:-99
Forward: 1 2 3 4 5
Backward: 5 4 3 2 1