指向 NULL 指针的指针在 C 中如何工作？

Question

我试图理解将两个排序列表合并为一个排序输出。从互联网上获得以下代码，理想情况下，

trav

的值必须是 mergedHead 的地址，但在这里，

trav

保存着

mergedHead

的值，并且

*mergedHead

的值等于

*trav

的值价值。谁能告诉我为什么会发生这种情况？另外请解释一下为什么这里的 mergedHead 的地址是 0。 此页面没有帮助。

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

typedef struct Node {
    int data;
    struct Node *next;
}Node;

Node *createNode(int data) {
    Node *newNode = (Node *) malloc(sizeof(Node));
    if(!newNode) {
        printf("mem-alloc failed for data: %d\n", data);
        exit(EXIT_FAILURE);
    }

    newNode->data = data;
    newNode->next = NULL;

    return newNode;
}

Node *mergeTwoSortedLists(Node *head1, Node *head2) {
    Node *mergedList = NULL;
    Node **trav = &mergedList;

    if(!head1 && !head2) {
        printf("Both the heads are NULL, cant merge\n");
        return NULL;
    }

    while (head1 && head2) {
        if(head1->data <= head2->data) {
            *trav = createNode(head1->data);
            printf("&trav: %p trav: %p, *trav: %p, **trav: %d, &mergedList: %p, mergedList: %p, *mergedList: %p\n", &trav, trav, *trav, **trav, &mergedList, mergedList, *mergedList);
            head1 = head1->next;
        }
        else {
            *trav = createNode(head2->data);
            printf("&trav: %p trav: %p, *trav: %p, **trav: %d, &mergedList: %p, mergedList: %p, *mergedList: %p\n", &trav, trav, *trav, **trav, &mergedList, mergedList, *mergedList);
            head2 = head2->next;
        }
        trav = &((*trav)->next);
    }

    /* insertion of the remaining list to the trav 
       or insert the list that is not null. */
    while (head1) {
        *trav = createNode(head1->data);
        printf("&trav: %p trav: %p, *trav: %p, **trav: %d, &mergedList: %p, mergedList: %p, *mergedList: %p\n", &trav, trav, *trav, **trav, &mergedList, mergedList, *mergedList);
        head1 = head1->next;
        trav = &((*trav)->next);
    }

    while (head2) {
        *trav = createNode(head2->data);
        printf("&trav: %p trav: %p, *trav: %p, **trav: %d, &mergedList: %p, mergedList: %p, *mergedList: %p\n", &trav, trav, *trav, **trav, &mergedList, mergedList, *mergedList);
        head2 = head2->next;
        trav = &((*trav)->next);
    }

    return mergedList;
}

void createSortedList(Node **head, int data) {
    Node *trav = NULL;
    Node *newNode = (Node *) malloc(sizeof(Node));

    if(!newNode) {
        printf("Memalloc failed - No memory allocated for data: %d\n", data);
        return;
    }

    newNode->data = data;

    if(!(*head) || (*head)->data >= data) {
        newNode->next = *head;
        *head = newNode;
        return;
    }

    trav = *head;

    while((trav->next) && (trav->next->data < data)) {
        trav = trav->next;
    }

    newNode->next = trav->next;
    trav->next = newNode;

    return;
}

void printList(Node *head) {
    if(!head) {
        printf("Node is empty\n");
        return;
    }
    while(head) {
        printf("%d\t", head->data);
        head = head->next;
    }
    printf("\n");
    return;
}

int main() {
    Node *head = NULL;
    Node *head2 = NULL;
    Node *head3 = NULL;
    createSortedList(&head, 3);
    createSortedList(&head, 5);
    createSortedList(&head, 7);
    createSortedList(&head, 3);
    createSortedList(&head, 4);
    createSortedList(&head, 9);
    printList(head);
    createSortedList(&head2, 10);
    createSortedList(&head2, 5);
    createSortedList(&head2, 9);
    createSortedList(&head2, 4);
    printList(head2);
    head3 = mergeTwoSortedLists(head, head2);
    printList(head3);
    return 0;
}

输出：

PS C:\Users\42410\Desktop\C\VSCode\LinkedList> .\a.exe
3       3       4       5       7       9
4       5       9       10
&trav: 0061FED8 trav: 0061FEDC, *trav: 007C0E28, **trav: 3, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E2C, *trav: 007C0E38, **trav: 3, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E3C, *trav: 007C0E48, **trav: 4, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E4C, *trav: 007C0E58, **trav: 4, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E5C, *trav: 007C0E68, **trav: 5, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E6C, *trav: 007C0E78, **trav: 5, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E7C, *trav: 007C0E88, **trav: 7, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0E8C, *trav: 007C04B0, **trav: 9, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C04B4, *trav: 007C0F20, **trav: 9, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
&trav: 0061FED8 trav: 007C0F24, *trav: 007C0EE0, **trav: 10, &mergedList: 00000000, mergedList: 0061FEDC, *mergedList: 007C0E28
3       3       4       4       5       5       7       9       9       10
PS C:\Users\42410\Desktop\C\VSCode\LinkedList>

Answer 1

您的打印声明是错误的。

如果您没有从编译器收到任何警告，则需要提高警告级别。例如，对于

gcc

使用选项：

-Wall -Wextra -Werror -pedantic

打印指针（即

%p

），您需要将指针强制转换为空指针。喜欢：

int x;
int* p = &x;
printf("P equals %p\n", (void*)p);
                        ^^^^^^^
                        cast to void*

更糟糕的是

**trav

属于

Node

类型，但您使用

%d

打印它。您可能想打印

(**trav).data

。

和

*mergedList

类似。再次是

Node

，但这次您使用

%p

打印它。也许您想要

%d

和

(*mergedList).data

。