c - 指针 链表 C 编程

Question

我不明白为什么这不起作用......例如我有这个。

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

static int length(struct node* head) {
   Does Stuff

};

void main() (  
   int i; 
   struct node* head;
   i = length(head);
);

但是代码不想工作...我得到了错误的输出。我试图将指针发送到我的函数，以便它们可以访问我 malloc 的数据。我将在下面发布完整的代码:

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

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

static int length(struct node* head);
static void push(struct node* head, int data);
static int pop(struct node* head);
static void appendNode(struct node* head, int data);
static struct node *copyList(struct node* head);
static void printList(struct node* head);



/************************************************************
 length - return length of a list
 ************************************************************/
int length(struct node* head) {
    int count = 0;
    struct node* current = NULL;

    current = head;
    while (current != NULL) {
        current = current->next;
        ++count;
    }

    return count;
}


/************************************************************
 push - add new node at beginning of list
 ************************************************************/
void push(struct node* head, int data) {
    struct node* new_ptr = NULL;

    new_ptr = (struct node*)malloc(sizeof(struct node));
    new_ptr->data = data;
    new_ptr->next = head;

    head = new_ptr;
}

/************************************************************
 pop - delete node at beginning of non-empty list and return its data
 ************************************************************/
int pop(struct node* head) {
    int val = 0;
    struct node* temp = NULL;

    if (head != NULL) {
        val = head->data;
        temp = head->next;
        free(head);
        head = temp;
    }

    return(val);
}

/************************************************************
 appendNode - add new node at end of list
 ************************************************************/
void appendNode(struct node* head, int data) {
    struct node* current = NULL;
    struct node* previous = NULL;
    struct node* new_ptr = NULL;

    current = head;
    previous = current;
    while (current != NULL) {
        previous = current;
        current = current->next;
    }

    new_ptr = (struct node*)malloc(sizeof(struct node));
    new_ptr->data = data;
    new_ptr->next = NULL;

    previous = new_ptr;

}

/************************************************************
 copyList - return new copy of list
 ************************************************************/
struct node* copyList(struct node* head) {
    struct node* copy = NULL;
    struct node* current = NULL;
    struct node* new_ptr = NULL;

    /* Copy current head to copy */
    current = head;
    while (current != NULL) {
        appendNode(copy, current->data);
        current = current->next;
    }

    return copy;
}


/************************************************************
 printList - print linked list as "List: < 2, 5, 6 >" (example)
 ************************************************************/
void printList(struct node* head) {
    struct node* current = NULL;

    printf("List: < ");

    current = head;
    if (current == NULL)
        printf("none ");

    while (current != NULL) {
        printf("%d", current->data);
        current = current->next;
        if (current != NULL)
            printf(", ");
    }

    printf(" >\n");
}

void main() {
    int i;                      // index used for loops
    struct node *list_a;        // a new list
    struct node *list_a_copy;   // copy of list
    list_a = NULL;                // initialize empty list
    list_a_copy = NULL;           // initialize empy list


    // test push
    for (i = 0; i < 4; ++i)
        push(list_a, i);

    // test length
    printf("Length of list = %d\n", length(list_a));

    // test print head list
    printf("head:\n");
    printList(list_a);

    // test append node
    for (i = 4; i < 8; ++i)
        appendNode(list_a, i);

    // test print head list
    printf("head(append):\n");
    printList(list_a);

    // make a copy of list
    list_a_copy = copyList(list_a);

    // test pop head list
    for (i = 0; i < 4; ++i)
        printf("%d popped\n", pop(list_a));

    // test print copy list
    printf("head copy:\n");
    printList(list_a_copy);

    // test pop copy list
    for (i = 0; i < 4; ++i)
        printf("%d popped\n", pop(list_a_copy));

}

谢谢你的帮助。我仍在学习这些 C 指针，而且我知道我已经很接近了。

干杯

Answer 1

我研究了函数push():

void push(struct node* head, int data) {
    struct node* new_ptr = NULL;

    new_ptr = (struct node*)malloc(sizeof(struct node));
    new_ptr->data = data;
    new_ptr->next = head;

    head = new_ptr;
}

您分配head = new_ptr;的方式是错误的。只有这样做，head 才会在函数内部生效，head 不会指向您在调用push() 后分配的内存。所以你需要修复你的push()函数:

void push(struct node **head, int data) {
    if ((*head) == null)
        (*head) = (struct node*)malloc(sizeof(struct node));
    (*head)->data = data;
    (*head)->next = head;
}