访问结构成员时出现分段错误

Question

I've been learning C and am having problems using linked lists. When looping over a pointer to a linked list I run into segmentation faults and I'm not sure why.

Looking at similar questions the suggestion is to allocate the memory, but I find this answer confusing. Do you have to use heap memory for linked lists, and if so why?

Here is my code:

#include <stdio.h>

typedef struct Node {
  char *name;
  struct Node *next;
} Node;

typedef struct Thing {
  Node *node;
} Thing;

Thing make_thing()
{
  Thing t = {
    .node = NULL
  };
  return t;
}

Thing * add_node(Thing *t, char *name)
{
  Node node = {
    .name = name,
    .next = t->node
  };

  t->node = &node;

  return t;
}

void print_nodes(Thing *t)
{
  Node *n = t->node;

  while(n != NULL) {
    printf("Node: %s\n", n->name);
    n = n->next;
  }
}

int main()
{
  printf("Start\n");

  Thing t = make_thing();
  add_node(&t, "one");

  printf("First %s\n", t.node->name);

  print_nodes(&t);

  return 0;
}

Answer 1

You are using objects with automatic storage out of their scope:

Node node = {
  .name = name,
  .next = t->node
};

t->node = &node;

return t;

Here you leak the pointer &node , which is invalid (out of scope) after the return, to the caller and use it here:

 printf("First %s\n", t.node->name);

You have to allocate memory by using malloc() for your Node structure.

Example:

 Node *node = malloc(sizeof *node);
 node->name = name;
 node->next = t->node;
 t->node = node;

 return t;

You have to care about freeing the memory when it is no longer used to prevent memory leaks.