指针在链表c实现中保持为空

Question

Sorry to bother you guys but I'm having difficulty understanding why my linked list implementation stays null. I'm trying to append elements to the beginning of the list. It's been a long time since I programmed in C and I simply can't spot the bug after hours of checking. Is it because I assign e to list and then free e??

#include <stdio.h>
#include <stdlib.h>
struct element{
  struct element *next;
  int i;
};
typedef struct element element;

void add(element *list, int n){
  element *e;
  e = malloc(sizeof(element));
  e->i = n;
  e->next = list;
  list = e;
  free(e);

}

void display(element *list){
  element *p;
  p = list;
  while(p!=NULL){
    printf(" %d\n",p->i );
    p=p->next;
  }

}

int main(void){
  element *list,*tail,*e, *p;
  //list = (element *)malloc(sizeof(element));
  //list->next = 0;
  list=NULL;
  add(list, 1);
  add(list,2);  
  add(list, 3); 
  display(list);
  //printf("%d\n",list->next);
  free(list);

  return 0;
}

Okay, thanks to bubbles for suggesting extra level of indirection. I'm such a noob... why do I only need only ONE level of indirection to add element to the end of the list as shown below?

#include <stdio.h>
#include <stdlib.h>
struct element{
  struct element *next;
  int i;
};
typedef struct element element;

void addB(element **list,int n){
    element *e = malloc(sizeof(element));
    e->i = n;
    e->next = *list;
    *list = e;
    //printf("%d\n", list->i);
}

void addE(element *list, int n){
  element *e = malloc(sizeof(element));
  e->i = n;
  e->next = NULL;
  element *p;
  p = list;
  if(!(list)){
    list = e;
  }else{
      while(p->next){
        p=p->next;
      }
      p->next = e;
  }

}

void display(element *list){
  element *p;
  p = list;
  while(p!=NULL){
    printf(" %d\n",p->i );
    p=p->next;
  }

}

void freelist(element *list){
  if(!(list->next)){
    free(list);
  }else{
    freelist(list->next);
  }

}

int main(void){
  element *list;
  list = NULL;
  addB(&list,1);
  addB(&list,2);
  addB(&list,3);
  addE(list,4);
  display(list);
  freelist(list);

  return 0;
}

Answer 1

You are freeing e in the same routine you are creating it. Instead you need to remove that free. For freeing the entire list, you will need a more substantive routine that traverses the list, freeing each element that was malloc'ed using the add routine.

Answer 2

The problem is that you are freeing "e". even after you are assigning e to list, e still points to the same memory location. Freeing e will simply make the allocated memory back to the part of free heap. Also the statement list = e will not have any effect, because you are altering a local copy of list. Modify the value of list variable you will have to define your add function something like

void add(element **list, int n){
  element *e;
  e = malloc(sizeof(element));
  e->i = n;
  e->next = *list;
  *list = e;

}


int main(void){
  element *list,*tail,*e, *p;
  //list = (element *)malloc(sizeof(element));
  //list->next = 0;
  list=NULL;
  add(&list, 1);
  add(&list,2);  
  add(&list, 3); 
  display(list);
  //printf("%d\n",list->next);
  free(list); //Should be replaced with a loop.

  return 0;
}

The way you have freed the list will free only the last node of it. You will have to write a loop to free the complete list.