将链接列表中的数据写入C中的txt文件

时间：2016-11-14 15:57:22

标签： c linked-list

我创建了这个程序，首先询问您拥有多少宠物，然后将每个宠物的名称和年龄存储在一个结构中（所有使用链接列表）。

我的问题是：我尝试使用过程writeToFile()将数据写入.txt文件，但在执行时，.txt文件不包含任何数据。我不明白为什么？

这是我的代码：

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

struct Node {
    char *name;
    int age;
    struct Node *next;
};

struct Node * petRecord;
struct Node * newRecord;

void printPetRecord()
{  
    while(petRecord != NULL)
    {
        printf("Name of Pet: %s\n", petRecord->name);
        printf("Age of Pet: %d\n", petRecord->age);
        petRecord = petRecord->next;
    }
}

void writeToFile()
{
    FILE * fptr;
    fptr = fopen("petnames.txt", "w");

    if(fptr==NULL)
    {
        printf("Error\n");
    }

    else
    {
        while(petRecord != NULL)
        {
            fprintf(fptr, "\nPet Name: %s\nAge: %d\n", petRecord->name, petRecord->age);
            petRecord = petRecord->next;
        }
    }

    fclose(fptr);
    }

int main()
{ 
    int count, i;
    printf("How many pets do you have? ");
    scanf("%d", &count);

    for(i=0; i<count; i++)
    {
        if(i==0)
        {
            petRecord = malloc(sizeof(struct Node));
            newRecord = petRecord;
        }
        else
        {
            newRecord->next = malloc(sizeof(struct Node));
            newRecord = newRecord->next;
        }
        newRecord->name = malloc(50*sizeof(char));
        printf("Name of Pet: ");
        scanf("%s", newRecord->name);
        printf("Age of Pet: ");
        scanf("%d", &newRecord->age);
    }
    newRecord->next = NULL;
    printf("\n\n");
    printPetRecord();
    writeToFile();
}

2 个答案:

答案 0 :(得分：3)

您的函数printPetRecord（）将指针设置为null。

在printPetRecord（）里面做这样的事情：

struct Node * iterator = petRecord;

然后使用迭代器迭代。

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

struct Node {
    char *name;
    int age;
    struct Node *next;
};

struct Node * petRecord;
struct Node * newRecord;

void printPetRecord()
{
    struct Node * iterator = petRecord;
    while(iterator != NULL)
    {
        printf("Name of Pet: %s\n", iterator->name);
        printf("Age of Pet: %d\n", iterator->age);
        iterator=iterator->next;
    }
}

void writeToFile()
{
    FILE * fptr;
    fptr = fopen("petnames.txt", "w");
    struct Node * iterator = petRecord;

    if(fptr==NULL)
    {
        printf("Error\n");
    }

    else
    {
        while(iterator!= NULL)
        {
            fprintf(fptr, "\nPet Name: %s\nAge: %d\n", iterator->name, iterator->age);
            iterator= iterator->next;
        }
    }

    fclose(fptr);
    }

int main()
{
    int count, i;
    printf("How many pets do you have? ");
    scanf("%d", &count);

    for(i=0; i<count; i++)
    {
        if(i==0)
        {
            petRecord = malloc(sizeof(struct Node));
            newRecord = petRecord;
        }
        else
        {
            newRecord->next = malloc(sizeof(struct Node));
            newRecord = newRecord->next;
        }
        newRecord->name = malloc(50*sizeof(char));
        printf("Name of Pet: ");
        scanf("%s", newRecord->name);
        printf("Age of Pet: ");
        scanf("%d", &newRecord->age);
    }
    newRecord->next = NULL;
    printf("\n\n");
    printPetRecord();
    writeToFile();
}

执行：

> gcc -o main main.c
> ./main
How many pets do you have? 2
Name of Pet: a
Age of Pet: 2
Name of Pet: b
Age of Pet: 3


Name of Pet: a
Age of Pet: 2
Name of Pet: b
Age of Pet: 3
> cat petnames.txt

Pet Name: a
Age: 2

Pet Name: b
Age: 3

答案 1 :(得分：1)

3年7个月后，我以某种方式解决了将linked list的数据写入.txt文件的问题时遇到了这个问题。我来这里是为了寻找解决问题的方法，但是这个问题没有得到任何答案。因此，在这里我将尝试回答这个问题。

  1. 程序中不需要全局变量。
  2. 向程序中的函数发送一些参数，这会使您的任务容易得多。
  3. 在这些函数内部进行迭代时，通过执行此head和在主petRecord = petRecord->next内部，您正在丢失newRecord = newRecord->next结构变量。这是由于全局变量而发生的。因此，为避免这种情况，您需要每次在被调用函数内部声明local类型的struct Node指针变量。
  4. 请勿执行以下操作：newRecord->name = malloc(50*sizeof(char));，如果您知道char数组的确切大小而不是这样做，请将struct模板修改为char name[50]。比使用指针和动态分配要容易得多，因为在后一种情况下，您需要格外小心和谨慎。
  5. 打开文件，其中将在data函数中写入main()，并将其发送到函数writeToFile()。
  6. 处理strings时应格外小心。
  7. 值得在分配内存或打开文件时检查错误。因此，每次在程序中执行任何上述操作时，最好检查一下在操作过程中发生的任何错误。

您为什么不尝试此程序： 

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

typedef struct Node {
    char name[50];
    int age;
    struct Node *next;
}Node;

// user-defined functions
void printPetRecord(Node *head);
void writeToFile(FILE *fptr, Node *head);

// main()
int main(void)
{
    int count, i;
    Node *petRecord, *newRecord;
    FILE *fp;

    if( (petRecord = malloc(sizeof(Node))) == NULL )
    {
        fprintf(stderr, "Unable to allocate memory.\n");
        exit(2);
    }
    newRecord = petRecord;
    printf("How many pets do you have? ");
    scanf("%d", &count);

    for(i = 0; i < count; i++)
    {
        printf("Name of Pet: ");
        scanf("%50s", newRecord->name);
        printf("Age of Pet: ");
        scanf("%d", &newRecord->age);
        if(i == count-1)
        {
            newRecord->next = NULL;
        }
        else
        {
            if( (newRecord->next = malloc(sizeof(Node))) == NULL)
            {
                fprintf(stderr, "Memory Unavailable.\n");
                exit(3);
            }
        }
        newRecord = newRecord->next;
    }
    printf("\n\n");
    // Modified arguments
    printPetRecord(petRecord);

    // Open file before sending to writeToFile
    if(!(fp = fopen("petname.txt", "w")))
    {
        fprintf(stderr, "Unable to open file \"petname.txt\"\n");
        exit(1);
    }
    // Modified arguments
    writeToFile(fp, petRecord);
    
    fclose(fp);
    return 0;
}

// function to print linked_list
void printPetRecord(Node *head)
{
    if(head->next != NULL)
    {
        printf("Name of Pet: %s\nAge of Pet: %d\n", head->name, head->age);
        printPetRecord(head->next);
    }
    else
        printf("Name of Pet: %s\nAge of Pet: %d\n", head->name, head->age);
}

// function to print list to file
void writeToFile(FILE *fptr, Node *head)
{
    if(head->next != NULL)
    {
        fprintf(fptr, "\nPet Name: %s\nAge: %d\n\n", head->name, head->age);
        writeToFile(fptr, head->next);
    }
    else
        fprintf(fptr, "\nPet Name: %s\nAge: %d\n\n", head->name, head->age);
}

输出（在控制台上）： 

How many pets do you have? 3
Name of Pet: Tommy
Age of Pet: 3
Name of Pet: Julia
Age of Pet: 4
Name of Pet: Hedgehog
Age of Pet: 5


Name of Pet: Tommy
Age of Pet: 3
Name of Pet: Julia
Age of Pet: 4
Name of Pet: Hedgehog
Age of Pet: 5

输出（在petname.txt文件内部）： 


Pet Name: Tommy
Age: 3


Pet Name: Julia
Age: 4


Pet Name: Hedgehog
Age: 5
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