使用 scanf 将标准输入解析为链表

Question

我想将一个由 3 个分号分隔的标准输入（例如：test1:127:completed）放入链表的一个节点中，该节点包含 3 个属性。当我运行这给了我一个段错误，我是否在第 96 行不正确地使用了 scanf？

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define BUFFERSIZE 1024

//for the linked list
struct node{
    char *fileName;
    int *lineNumber;
    char *filler;
    struct node *next;
};

int main (int argc, char *argv[])
{
    //Create the linked list of the file location and the line number
    char tmpstring[BUFFERSIZE];
    /* This won't change, or we would lose the list in memory */
    struct node *head = NULL;
    //tail
    struct node *tail = NULL;
    /* This will point to each node as it traverses the list */
    struct node *ptr;
    //new node pointer
    struct node *newnode;

    while (fgets(tmpstring, 128, stdin) == tmpstring)
    {
        newnode->next = NULL;
        // First insertion is a special case.
        if (tail == NULL)
            head = tail = newnode;
        else {
            tail->next = newnode;
            tail = newnode;
        }
    } // while

    for (ptr = head; ptr != NULL; ptr = ptr->next)
        //THIS IS THE PROBLEM                    scanf(" %s:%d:%s", ptr -> fileName, ptr -> lineNumber, ptr -> filler);
        printf("%s", ptr->fileName);

    //get how many nodes are in the linked list
    struct node *tmp = head;
    int count = 0;
    while(tmp!=NULL)
    {
        count++;
        tmp = tmp->next;
    }
    //end list creation

    //getGrepOutput();
    printf("%d\n", count);
}

//new code added
ptr -> fileName = (char*) malloc(50);
ptr -> lineNumber = (int*) malloc(50);
ptr -> filler = (char*) malloc(50);

Answer 1

scanf(" %s:%d:%s", ptr -> fileName, ptr -> lineNumber, ptr -> filler);

您需要为 ptr -> fileName 分配内存，这只是您程序中的悬垂指针。 lineNumber 和 filler 成员也一样。

使用 scanf 将标准输入解析为链表

Parsing std input into a linked list using scanf

c

scanf