c - C中遍历和打印单链表时出错

Question

迷失了菜鸟，尝试逐行读取文件，即“一”、“二”、“三”并将其添加到有序链表中(我相信我已经工作了)。但是，我无法弄清楚 traverse_and_print 列表函数的语法/逻辑(并且不理解 *(转到此处并获取值)、&(获取地址)和 -> 。我主要工作的代码位于repl.it 位于 https://repl.it/@MichaelB4/DeafeningTreasuredMathematics

// A complete working C program to demonstrate all insertion methods 
// from https://www.geeksforgeeks.org/linked-list-set-2- inserting-a-node/

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

// Create structure for a linked list node 

struct Node 
{ 
    const char *data; 
    struct Node *next; 
}; 

struct Node *head;

// Given a reference (pointer to pointer) to the head of a list and a char*, appends a new node at the end
void append(struct Node** head_ref, const char *new_data) 
{ 
    // 1. allocate node
    struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); 

    struct Node *last = *head_ref;  // used in step 5

    // 2. put in the data
    new_node-> data  = new_data; 

    // 3. Set new_node.next to Null, as it will be inserted at tail of list 
    new_node -> next = NULL;

    // 4. If the Linked List is empty, then make the new node as head
    if (*head_ref == NULL) 
    { 
        *head_ref = new_node;
        //printf("head\n");
        printf("%s", new_node -> data); 
        return; 
    } 

    // 5. Else traverse till the last node

    while (last -> next != NULL)  
        last = last -> next;
        printf("%s", new_node -> data); 

    // 6. Change the next of last node, have last node point to one just inserted, the new node at the end of the list, tail 

    last -> next = new_node; 
    return; 
} 

void traverse_and_printList(head){
    struct Node* current = (struct Node*) malloc(sizeof(struct Node));
    current = head;
    while (head -> next != NULL)
        printf("%s", current -> data);
        current -> next = current;

}

/* Driver program to test above functions*/
int main() 
{ 
    // set up a file point to File to be opened
    FILE* fp;
    // holds contents of each line/word 
    char buffer[255]; 

    fp = fopen("words.txt", "r");
    if (fp == NULL)
    {
        fprintf(stderr, "Could not open infile"); 
        return 2;
    }

    /* create an empty node */
    struct Node* head = NULL; 

    int counter = 0;
    char *head_value[255]; 

    while(fgets(buffer, 255, (FILE*) fp)){
        //printf("%s", buffer);
        append(&head, buffer);
    }

    fclose(fp);
    traverse_and_printList(head);
    printf("\n");
    return 0; 
}

Answer 1

您对 current 指针的用途感到困惑。

首先，您不需要为其分配内存。您并不是想存储任何新内容。 current 指针只是一个帮助您在列表中的项目上移动的值。

其次，您不应修改列表的数据。 current->next = current 行是伪造的。它修改列表并创建一个循环。不好。

第三，您的缩进表明您的 while 循环包含两个单独的语句，但没有 block 作用域(即 { ... }) 在他们周围。因此，只有第一个语句将成为循环的一部分。

最后，关于风格的一点。请不要在 -> 周围添加空格。虽然编译器不在乎，但它会使您的代码非常难以人类阅读。

正确遍历列表就像这样简单:

for(struct Node* current = head; current != NULL; current = current->next)
{
    printf("%s\n", current->data);
}