反向打印雙向鏈表

Question

我有一個雙向鏈表，可以從上到下打印，現在我正嘗試從下到上打印它。

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

//defines the struct UserData
typedef struct
{
    int importance;
    char taskName[80];
}UserData, *UserDataPtr;

//Defines a node
typedef struct node {
    UserData Data;
    struct node *next;
    struct node *prev;
    } Node, *NodePtr;

NodePtr makeNode(UserData);

//Declare function printList
void printList(NodePtr);

void printListRev(NodePtr);

int main()
{
   UserData info;
   NodePtr top, ptr, last, temp;



   top = NULL;

    FILE *filein=fopen("Data.txt", "r");
    if (filein == NULL) {
        printf("Error opening file, exiting program.\n");
        exit(0);
    }

    while(fscanf(filein, "%d%s",&info.importance, info.taskName)==2)
        {
            ptr=makeNode(info);
            if (top == NULL) top = ptr;
            else last -> next = ptr;
            last = ptr;
        }//end while loop

    printList(top);


    printListRev(last);
   }//end Main


//printList is a function that prints each node as long as it isn't NULL. Once it reaches NULL it terminates, signifying the end of the list.
void printList(NodePtr ptr) {
    while (ptr != NULL) { //as long as there's a node
            printf("%d %s\n", ptr -> Data.importance, ptr -> Data.taskName);
            ptr = ptr -> next; //go on to the next node

        }
    if (ptr == NULL) {

        printf("Last node data printed moving forward.\n");
    }

    } //end printList


void printListRev(NodePtr ptr) {
    while(ptr != NULL){
        printf("%d %s\n", ptr -> Data.importance, ptr -> Data.taskName);
        ptr = ptr -> prev;


      }
    }//end printListRev

//Define function makeNode. Allocates storage for node, stores integer given to it, and returns a pointer to the new node. Also sets next field to NULL
NodePtr makeNode(UserData info) {
    NodePtr ptr = (NodePtr) malloc(sizeof (Node));
    ptr -> Data = info;
    ptr -> next = NULL;
    ptr -> prev = NULL;
    return ptr;
} //End makeNode

這是輸出：

1 task1
2 task2A
3 task3A
2 task2B
4 task4A
4 task4B
3 task3B
Last node data printed moving forward.
3 task3B

而且我不知道為什么它不會反向打印完整列表。 反向打印時僅打印一項。

直到“最后一個節點數據已打印”消息為止，所有內容都是正確的。 是的，這有點混亂，我是C的新手，我需要清理我的注釋等。 道歉。

有人可以幫忙嗎？

Answer 1

將節點插入列表時，您忘記將prev字段設置為適當的值。 解決方法很簡單：將last初始化為NULL ，然后設置ptr->prev = last; 在ptr = makeNode(info);行之后ptr = makeNode(info);

順便說一句， temp未使用。

Answer 2

在讀取階段，您設置了last->next ，但從未將任何prev成員設置為NULL以外的任何其他NULL 。

你可以看到這一點，如果你修改printList()代碼打印prev和next成員太-使用%p轉換符。

例如：

void printList(NodePtr ptr)
{
    while (ptr != NULL)
    {
        printf("%d %s (N = %p, P = %p)\n", ptr->Data.importance, ptr->Data.taskName,
               (void *)ptr->next, (void *)ptr->prev);
        ptr = ptr->next;
    }
    if (ptr == NULL)
    {
        printf("Last node data printed moving forward.\n");
    }
}

運行時，將產生（對我來說，在我的Mac上）：

1 task1 (N = 0x7ff6a94032e0, P = 0x0)
2 task2A (N = 0x7ff6a9403370, P = 0x0)
3 task3A (N = 0x7ff6a94033e0, P = 0x0)
2 task2B (N = 0x7ff6a9403450, P = 0x0)
4 task4A (N = 0x7ff6a94034c0, P = 0x0)
4 task4B (N = 0x7ff6a9403530, P = 0x0)
3 task3B (N = 0x0, P = 0x0)
Last node data printed moving forward.
3 task3B

如您所見，沒有反向鏈接，因此在打印一個元素（無論您指向哪個元素）后，反向打印都會停止。

^{請注意，在編寫C時，不應在兩個點之間使用空格. 或箭頭->運算符。 它們之間的綁定非常緊密，不應使用空格（因為從語法上講，所有空格都是合法的）。 如果使用這種非正統的布局，則代碼的可讀性將大大降低。}

掃描代碼中的修復很簡單：

    while (fscanf(filein, "%d%s", &info.importance, info.taskName) == 2)
    {
        ptr = makeNode(info);
        if (top == NULL)
            top = ptr;
        else
            last->next = ptr;
        ptr->prev = last;
        last = ptr;
    }

我還初始化了last = NULL; 在循環開始之前； 當您使用它來設置上一個指針時，這一點至關重要。 盡管GCC抱怨我的默認編譯選項“可能未初始化使用”，但您以前可以省略它。 實際上，並不是未初始化就使用了它，但是可以理解，編譯器（GCC 6.2.0）是可以理解的。

進行此更改后，輸出為：

1 task1 (N = 0x7fa4b1602a10, P = 0x0)
2 task2A (N = 0x7fa4b1602aa0, P = 0x7fa4b16029a0)
3 task3A (N = 0x7fa4b1602b10, P = 0x7fa4b1602a10)
2 task2B (N = 0x7fa4b1602b80, P = 0x7fa4b1602aa0)
4 task4A (N = 0x7fa4b1602bf0, P = 0x7fa4b1602b10)
4 task4B (N = 0x7fa4b1602c60, P = 0x7fa4b1602b80)
3 task3B (N = 0x0, P = 0x7fa4b1602bf0)
Last node data printed moving forward.
3 task3B
4 task4B
4 task4A
2 task2B
3 task3A
2 task2A
1 task1

您也可以打印該節點的地址。 這樣可以更輕松地跟蹤每個列表指針都指向正確的位置：

void printList(NodePtr ptr)
{
    while (ptr != NULL)
    {
        printf("%d %s (C = %p, N = %p, P = %p)\n", ptr->Data.importance, ptr->Data.taskName,
               (void *)ptr, (void *)ptr->next, (void *)ptr->prev);
        ptr = ptr->next;
    }
    printf("Last node data printed moving forward.\n");
}

void printListRev(NodePtr ptr)
{
    while (ptr != NULL)
    {
        printf("%d %s (C = %p, N = %p, P = %p)\n", ptr->Data.importance, ptr->Data.taskName,
               (void *)ptr, (void *)ptr->next, (void *)ptr->prev);
        ptr = ptr->prev;
    }
    printf("Last node data printed moving backward.\n");
}

生產：

1 task1 (C = 0x7fd301c03270, N = 0x7fd301c032e0, P = 0x0)
2 task2A (C = 0x7fd301c032e0, N = 0x7fd301c03370, P = 0x7fd301c03270)
3 task3A (C = 0x7fd301c03370, N = 0x7fd301c033e0, P = 0x7fd301c032e0)
2 task2B (C = 0x7fd301c033e0, N = 0x7fd301c03450, P = 0x7fd301c03370)
4 task4A (C = 0x7fd301c03450, N = 0x7fd301c034c0, P = 0x7fd301c033e0)
4 task4B (C = 0x7fd301c034c0, N = 0x7fd301c03530, P = 0x7fd301c03450)
3 task3B (C = 0x7fd301c03530, N = 0x0, P = 0x7fd301c034c0)
Last node data printed moving forward.
3 task3B (C = 0x7fd301c03530, N = 0x0, P = 0x7fd301c034c0)
4 task4B (C = 0x7fd301c034c0, N = 0x7fd301c03530, P = 0x7fd301c03450)
4 task4A (C = 0x7fd301c03450, N = 0x7fd301c034c0, P = 0x7fd301c033e0)
2 task2B (C = 0x7fd301c033e0, N = 0x7fd301c03450, P = 0x7fd301c03370)
3 task3A (C = 0x7fd301c03370, N = 0x7fd301c033e0, P = 0x7fd301c032e0)
2 task2A (C = 0x7fd301c032e0, N = 0x7fd301c03370, P = 0x7fd301c03270)
1 task1 (C = 0x7fd301c03270, N = 0x7fd301c032e0, P = 0x0)
Last node data printed moving backward.

Answer 3

您忘記了建立prev鏈接。

else last -> next = ptr;

應該

else {
    ptr->prev = last;
    last -> next = ptr;
}