c - 链表，一个元素丢失

Question

我正在尝试将一个节点插入链表的末尾，一切正常，但是，当涉及到最后一个元素时，它会打印 null。 else block 中一定有问题，我在想当我查找最后一项时它可能指向 NULL 而不是具有 next< 的元素 作为 NULL

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


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

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


struct person* insert_end(struct person *ptr, char *name, int age)
{
  //The new location for the new person
  struct person* newPer = malloc(sizeof(struct person));
    if (newPer == NULL) {
        printf("something went wrong allocating the memory\n");
        exit(1);
    }

  newPer -> name = name;
  newPer -> age = age;
  //Make its next points previous element
  if (ptr->next == NULL) {
    newPer -> next = ptr;


    return newPer;
  }
  else {
    struct person *tmp = ptr;
    while(tmp -> next != NULL){
      tmp = tmp -> next;

    }
    printf("%s", tmp -> name);
    tmp -> next = newPer;


    //if(strcmp("Harriet",name)==0)
    return ptr;
  }
  //return the new address so that it becomes the new HEAD of the linked list
    return newPer;
}


int main(int argc, char **argv)
{
    //This is the head of the list
    struct person *HEAD = NULL;
    HEAD = malloc(sizeof(struct person));
    if (HEAD == NULL) {
        printf("something went wrong allocating the memory");
        exit(1);
    }


    int i;
  //insert a new person and make HEAD points to it, so that HEADS
  //will be pointing to the last element added to the linked list.
  for (i = 0; i < HOW_MANY; i++){
        HEAD = insert_end (HEAD, names[i], ages[i]);
    }

  struct person* tmp = HEAD;
  //We can use the member name(pointer) as the condition, if we did then extra
  //unwanted elements added to the linked list by accident won't be printed
  while (tmp != NULL){
    if (tmp -> next != NULL)
      printf("The name is %s, the age is %d years\n", tmp->name, tmp->age);

    //store the pointer in a tmp so than we can access the next pointer then
    //free tmp
    struct person* prevtmp = tmp;
    tmp = tmp -> next;
    free(prevtmp);
  }
}

输出是

The name is Simon, the age is 22 years
The name is (null), the age is 0 years
The name is Suzie, the age is 24 years
The name is Alfred, the age is 106 years
The name is Chip, the age is 6 years
The name is John, the age is 18 years
The name is Tim, the age is 32 years

Answer 1

只需将 if (tmp -> next != NULL) 更改为 if (tmp != NULL)。您忘记了列表中的最后一个节点不会有“下一个”节点。