头在c链表中不断变化答案

【问题标题】：Head constantly changing in c linked list头在c链表中不断变化
【发布时间】：2017-05-29 09:16:00
【问题描述】：

我正在尝试在 C 中实现一个简单的链表，但似乎没有正确添加第一个元素。这是我的代码：

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

typedef struct person{
    int phoneNumber;
    char* name;
    struct person *nextPers;
}person;

person* firstPerson=NULL;
person* lastPerson=NULL;

void addPerson (person* _person){
/*Adds a new person to the linked list after the last one added
*/  
if (firstPerson!=NULL){
    fprintf(stderr,"DEBUG(addPerson): First person is %s \n",firstPerson->name);
    fprintf(stderr,"DEBUG (addPerson): Last Person is (before adding the new one) %s \n",lastPerson->name);}
fprintf(stderr,"DEBUG: Adding person %s \n",_person->name);
_person->nextPers= NULL;
if (firstPerson==NULL){
    firstPerson = _person;
    lastPerson= _person;
    fprintf(stderr,"DEBUG: The head of the list is %s \n",firstPerson->name);   

}else{
    fprintf(stderr,"DEBUG: Last person (before adding the new one) %s \n",lastPerson->name);
    fprintf(stderr,"DEBUG (addPerson):Adding to the list %s \n",_person->name); 
    lastPerson->nextPers =_person;
    lastPerson=_person;
    fprintf(stderr,"DEBUG: Last person %s \n",lastPerson->name);
    }
}


int main(int argc, char* argv[]){
    char line[80],name[80];
    int number;
    setvbuf(stdout,(char*)malloc(sizeof(char)*80),_IOLBF,80);
    setvbuf(stdin,(char*)malloc(sizeof(char)*80),_IOLBF,80);

    for(;fgets(line,80,stdin);){
        if(!strcmp(line,"Finish\n"))
              break;

        sscanf(line,"%[^:]: %d",name,&number);
        /* Stores the person introduced from stdin as Name:phone */
        if(firstPerson!=NULL){
            fprintf(stderr,"DEBUG (Before storing new data aparently): First person is %s \n",firstPerson>name);    
        }
        person * newPerson= malloc(sizeof(person));

        newPerson->phoneNumber = number;    
        newPerson->name = name;

        fprintf(stderr,"DEBUG: Adding person %s \n",newPerson->name);
        fprintf(stderr,"DEBUG: phone number %d \n",newPerson->phoneNumber); 

        addPerson(newPerson);

        fprintf(stderr,"DEBUG: There is a new person on the list\n");                   

        }
}

预期的输出是：

Harvey:12345
DEBUG: Adding person Harvey 
DEBUG: phone number 12345 
DEBUG: Adding person Harvey 
DEBUG: The head of the list is Harvey 
DEBUG: There is a new person on the list
Adam:23456 
DEBUG (Before storing new data aparently): First person is Harvey 
DEBUG: Adding person Adam 
DEBUG: phone number 23456 
DEBUG(addPerson): First person is Harvey
DEBUG (addPerson): Last Person is (before adding the new one) Harvey 
DEBUG: Adding person Adam 
DEBUG: Last person (before adding the new one) Harvey
DEBUG (addPerson):Adding to the list Adam 
DEBUG: Last person Adam  
DEBUG: There is a new person on the list

但是，输出却是：

Harvey:12345
DEBUG: Adding person Harvey 
DEBUG: phone number 12345 
DEBUG: Adding person Harvey 
DEBUG: The head of the list is Harvey 
DEBUG: There is a new person on the list
Adam:23456
DEBUG (Before storing new data aparently): First person is (null) //Wrong, its Harvey
DEBUG: Adding person Adam 
DEBUG: phone number 23456 
DEBUG(addPerson): First person is Adam //Wrong, its Harvey
DEBUG (addPerson): Last Person is (before adding the new one) Adam //Wrong, its Harvey
DEBUG: Adding person Adam 
DEBUG: Last person (before adding the new one) Adam //Wrong its Harvey
DEBUG (addPerson):Adding to the list Adam 
DEBUG: Last person Adam 
DEBUG: There is a new person on the list

总结一下，主要思路是得到一个链表就是：Harvey->Adam->(next) 但是不知道哪里出错了。

【问题讨论】：

fprintf(stderr, "DEBUG (Before storing new data aparently): First person is %s \n", firstPerson>name); -> fprintf(stderr, "DEBUG (Before storing new data aparently): First person is %s \n", firstPerson->name);
并注意编译器警告：lastPerson->nextPers = _person;：这里_person 的类型是person*，而lastPerson->nextPers 的类型是struct proceso *。你搞混了。
您向我们展示了预期的和实际的输出，这很好，但您也应该展示输入。
最后但同样重要的是：学习如何使用调试器，而不是进行“printf 调试”。它的效率要高得多，而且您花在学习上的时间会很快得到回报。

标签： c linked-list head

【解决方案1】：

你的代码的链表部分是正确的，我知道这会让你发疯。您的代码中还有一些其他错误，与链表无关。

            fprintf(stderr,"DEBUG (Before storing new data aparently): First person is %s \n",firstPerson>name);

看到你写的是 firstPerson>name 而不是 firstPerson->name，这是进行布尔比较并返回零 - 这就是为什么你有一个“null”。

第二个错误在这里：

    newPerson->phoneNumber = number;    
    newPerson->name = name

您正在将 phoneNumber 设置为名称和号码数组的指针。因此，当您在这里进行第二次打印时：

DEBUG (addPerson): Last Person is (before adding the new one) Adam //Wrong, its Harvey

它实际上是在检索正确的值 - 因为您将 Harvey 覆盖到了 Adam。

解决方法很简单：将字符串复制到 newPerson 而不是将其指向名称。

newPerson->name = malloc(sizeof(char) * strlen(name) + 1); 
strcpy(newPerson->name, name);

这应该可以解决您的问题。

【讨论】：

【解决方案2】：

你犯了一些错误：

struct proceso *nextPers;

这个没意义，改成

person *nextPers;

然后保存在结构中的名称指向 name[80]（main() 中的 var），因此每次您的 this var 更改时，您的结构中的 var 也会更改。你必须把它复制到你的结构上；

memcpy(newPerson->name, name, 80);

并将您的结构更改为：

typedef struct person{
    int phoneNumber;
    char name[80];
    struct person *nextPers;
}person;

还记得在完成后释放列表中的所有节点

【讨论】：