在指针中需要一些帮助

时间:2016-10-26 03:16:29

标签: c data-structures linked-list

在此计划中,我尝试在功能tail中打印tail->nexttail->dataSortedMerge(struct node* a, struct node* b)值。我创建了一个链接列表,如5->10->15,其头部指针为#34; a"和2->3->20有头指针" b":

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

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


struct node* SortedMerge(struct node* a, struct node* b)
{
    /* a dummy first node to hang the result on */
    struct node dummy;

    /* tail points to the last result node */
    struct node* tail = &dummy;
    printf("tail %d \n",tail);
    printf("tail->next %d \n",tail->next);
    printf("tail->data %d \n",tail->data);
}

/* Function to insert a node at the beginging of the
linked list */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node =
        (struct node*) malloc(sizeof(struct node));

    /* put in the data */
    new_node->data = new_data;

    /* link the old list off the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

/* Drier program to test above functions*/
int main()
{
    struct node* res = NULL;
    struct node* a = NULL;
    struct node* b = NULL;
    push(&a,5); //some more like this (5->10->15)
    push(&b,2); //some more like this (2->3->20)
    res = SortedMerge(a, b);

    return 0;
}

我的输出是这样的。

tail -686550032 
tail->next 15585456 
tail->data 1

任何人都可以解释我这个。

1 个答案:

答案 0 :(得分:1)

正如Ari0nhh所述,为避免未定义的行为,您的SortedMerge函数必须使用%p来打印指针地址,如下所示:

printf("tail %p \n",tail);
printf("tail->next %p \n",tail->next);

导致类似

的内容
tail 0x7fff9c7f2f80
tail->next 0x7fff9c7f2fb8
tail->data 0

但是,如果您想要与输入数据进行交互,则应在函数中使用它们:

struct node* SortedMerge_a(struct node* a, struct node* b)
{
  printf("a %p \n",a);
  printf("a->next %p \n",a->next);
  printf("a->data %d \n",a->data);
}

给出

a 0x601010
a->next (nil)
a->data 5