Malloc cannot allocate memory for a structure - c

Question

I have the Segmentation fault (core dumped) error.
main.c
#include "header1.h"
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char** argv) {
struct t_list *list = NULL;
doFill(list);
printf("%s\n", list->name);
free(list);
return 0;
}
header1.h
#ifndef HEADER1_H
#define HEADER1_H
struct t_list {
char *name;
struct t_list *next;
};
void doFill(struct t_list *list);
#endif
worker.c
#include "header1.h"
#include <stdlib.h>
void doFill(struct t_list *list) {
list = (struct t_list *) malloc(sizeof(struct t_list));
char *tmp = "somename";
list->name = tmp;
list->next = NULL;
}
When I run this (gcc -g main.c worker.c -o test) I get (on the line with printf in main.c):
Segmentation fault (core dumped)
In gdb I see:
Temporary breakpoint 1, main (argc=1, argv=0x7fffffffddf8) at main.c:8
8 struct t_list *list = NULL;
(gdb) next
9 doFill(list);
(gdb) step
doFill (list=0x0) at worker.c:6
6 list = (struct t_list *) malloc(sizeof(struct t_list));
(gdb) p list
$1 = (struct t_list *) 0x0
(gdb) next
7 char *tmp = "somename";
(gdb) p list
$2 = (struct t_list *) 0x0
As you can see malloc in worker.c doesn't allocate memory for the list variable (the pointer before and after malloc points at 0x0).
If I move the code from the doFill procedure in main.c it works correctly:
main.c
#include "header1.h"
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char** argv) {
struct t_list *list;
list = (struct t_list *) malloc(sizeof(struct t_list));
char *tmp = "somename";
list->name = tmp;
list->next = NULL;
printf("%s\n", list->name);
free(list);
return 0;
}
$ gcc -g main.c -o test
$ ./test
somename
How is it possible? What do I do wrong?
gcc version 4.8.2 (Ubuntu 4.8.2-19ubuntu1)

Answer 1

You aren't receiving back the new value of list. In fact, passing list in is totally useless. Better to pass in the name for this node.
typedef struct t_list List;
List *newListNode(char *name) {
List *list = malloc(sizeof(*list));
if (!list) return NULL;
list->name = strdup(name);
if (!list->name) { free(list); return NULL; }
list->next = NULL;
return list;
}
char *strdup(char *src) { // if strdup doesn't already exist.
char *dst = malloc(strlen(src) + 1);
if (!dst) return NULL;
strcpy(dst, src);
return dst;
}
To add nodes to the front of the list:
List *listAdd(List *list, char *name) {
List *newnode = newListNode(name);
if (!newnode) return NULL;
if (list) newnode->next = list;
return newnode;
}
To delete the list, remember to delete the malloced strings:
void deleteList(List *list) {
for (List *next; list; list = next) {
next = list->next;
free(list->name);
free(list);
}
}

Answer 2

Parameters in C are passed by copy. The changes you make to list inside of doFill() are not propagated back to main(), which means that list is always NULL in main(). Try passing a pointer to pointer instead:
#include "header1.h"
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char** argv) {
struct t_list *list = NULL;
doFill(&list);
printf("%s\n", list->name);
free(list);
return 0;
}
And then change doFill() accordingly:
#include "header1.h"
#include <stdlib.h>
void doFill(struct t_list **list) {
*list = malloc(sizeof(**list));
char *tmp = "somename";
(*list)->name = tmp;
(*list)->next = NULL;
}