比较列表项，并在 c 中按顺序排序

Question

代码已编译，但是我的代码中存在逻辑错误。我想比较数组中的字符串，然后在列表中按顺序列出它们。我不知道如何在不使用索引的情况下比较列表项，以及如何将当前名称与下一个名称进行比较。任何人都可以帮忙吗？

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


/* these arrays are just used to give the parameters to 'insert',
   to create the 'people' array */
char names[][10]= {"Simon", "Suzie", "Alfred", "Chip", "John", "Tim",
      "Harriet"};
int ages[7]= {22, 24, 106, 6, 18, 32, 24};


/* declare your struct for a person here */
typedef struct Record{
    char *name;
    int age;
    struct Record *next;
}   Record;

//set the head pointer at the start of the list
Record *headptr = NULL;

int compare_people( Record *a, Record *b)
{
     return strcmp((*(Record *)a).name, (*(Record *)b).name);
}

static void insert (Record *p, char *s, int n) {

    /* create a new space for the new person */
    Record *ptr = ( Record *) malloc(sizeof(Record));

    /* check if it is succeeded  */ 
    if( ptr == NULL){  
        abort();
        printf("memory allocation fail"); 
        exit(1);  
    }else{
        printf("memory allocation to person  - %s - \n", s);      
    }

    //set the data for the new person
    ptr->name=s;
    ptr->age=n;
    ptr->next= NULL;

    //ptr= NULL; 
    //printf("%i", p->age);


    /*  do not compare when the list is empty*/
    if(headptr==NULL)
    {
        ptr->next=headptr;
        headptr=ptr;
        printf("ok1\n");

    }else{
        Record *tail = headptr;

        /* go through all the list */
        while(tail->next!=NULL)
        {     
            if(compare_people(ptr->name,tail->name)== 1){
            tail = tail->next;
        }else{
            tail->next=headptr;
            }
        }//while

        //tail->next=ptr;
    }  
}  

int main( int argc, char **argv) {

    /* declare the people array here */
    Record *p=headptr;
    headptr = NULL;

    //insert the members and age into the unusage array. 
    for (int i=0; i < 7; i++) {
         insert (p,names[i], ages[i]);
         /* do not dereference the pointer */
    }

    /* print out a line before printing the names and ages */
    printf("\n");

    //set the pointer at the start of the list 
    p = headptr;

    /* print the people array here*/
    for ( int i=0; i < 7; i++, p = p->next ) {
        printf("The name is: %s, the age is:%i\n", p->name, p->age);
    }


    /* This is the third loop for call free to release the memory allocated by malloc */
    /* the free()function deallocate the space pointed by ptr. */
    for( int i=0; i<7; i++){
        free(p->next);
    } 
}

Answer 1

这段代码看起来不对：

Record *tail =headptr;
/* go through all the list */
while(tail->next!=NULL)
{   
  if(compare_people(ptr->name,tail->name)== 1){
    tail = tail->next;
  } else {
    tail->next=headptr;
  }
} //while

如果你想在tail 之后插入一些东西，简单地设置tail->next = headptr 将会（a）泄漏tail 之后的任何东西，并且（b）把你的链表变成一个没有结束的循环。

如果你想将ptr 插入到你的列表中，你可能应该这样做

ptr->next = tail->next; 
tail->next = ptr; 

...然后跳出循环。

Answer 2

第一个主要问题在这里：

Record *tail =headptr;
/* go through all the list */
while(tail->next!=NULL) {
...

你永远不会进入这个while() 循环。在第一次迭代中，您这样做了：

ptr->next= NULL;  // setting the pointer's next pointer to NULL
...
headptr=ptr;     // have headptr point at what ptr is pointing to

这意味着headptr->next 将是NULL。然后在上面的代码 sn-p 中将tail 设置为headptr，因此tail->next 将是NULL，并且您将永远不会执行该循环。

第二个主要问题来了：

if(compare_people(ptr->name,tail->name)== 1){

您正在向该函数传递一个字符串（record->name 是一个字符串），但在函数本身中您已将其设置为：

int compare_people(Record *a, Record *b)

将记录（不是char *）作为输入。一旦你解决了第一个问题并真正使用了这个函数，这会杀了你。

Answer 3

您的代码包含许多错误。我无法检查您的代码和 cmets 的所有错误。我试图修复你的代码。

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

char names[][10]= {"Simon", "Suzie", "Alfred", "Chip", "John", "Tim",
      "Harriet"};
int ages[7]= {22, 24, 106, 6, 18, 32, 24};


/* declare your struct for a person here */
typedef struct Record{
  char *name;
  int age;
  struct Record *next;
}  Record;

//set the head pointer at the start of the list
Record *headptr = NULL;

int compare_people(char *a, char *b)
{

  return strcmp(a, b);
}

void insert (char *s, int n) {
     Record *t, *pnew, *prv;
     int i;
     prv=NULL;
     pnew=(Record *)malloc(sizeof(struct Record));
     if(pnew == NULL){  
        abort();
        printf("memory allocation fail"); 
        exit(1);  
    }else{
        printf("memory allocation to person  - %s - \n", s);      
    }
     pnew->name = s;
     pnew->age = n;
     pnew->next = NULL;
     if (headptr==NULL)
     {
        headptr = pnew;
        return;
     }
     for (t=headptr;t!=NULL;t=t->next) { // look for the right place to insert in order to get a tri list
         if (compare_people(s,t->name)<0) {        
            pnew->next=t;
            if (prv!=NULL)
               prv->next = pnew;
            else
               headptr=pnew;
            return;
         }
         prv=t;
     }
     prv->next=pnew;
     return;       
}

int main(int argc, char **argv) {

  Record *p, *q;
  int i;

  for (i=0; i < 7; i++) {
     insert (names[i], ages[i]);
  }

   printf("\n");

  for (p = headptr; p!=NULL; p = p->next) {
    printf("The name is: %s, the age is:%i\n", p->name, p->age);
  }


  /* To free your linked list: */
  p = headptr;
  while (p!=NULL){
    q = p;
    p = p->next;
    free(q);
  }
}

上述代码执行的输出：

linux$ ./test
memory allocation to person  - Simon - 
memory allocation to person  - Suzie - 
memory allocation to person  - Alfred - 
memory allocation to person  - Chip - 
memory allocation to person  - John - 
memory allocation to person  - Tim - 
memory allocation to person  - Harriet - 

The name is: Alfred, the age is:106
The name is: Chip, the age is:6
The name is: Harriet, the age is:24
The name is: John, the age is:18
The name is: Simon, the age is:22
The name is: Suzie, the age is:24
The name is: Tim, the age is:32

Answer 4

您可能必须使用双重链表（添加指向列表前一条记录的指针）。然后将更容易对列表的元素进行排序。希望对你有帮助。