Linked List value changes in gdb - c

Question

I have a C linked list that looks like this:
typedef struct Node {
struct Node *child;
void *value;
} Node;
typedef struct LinkedList {
Node *head;
} LinkedList;
To test that everything is working properly, I have a main program that reads from a file, line by line, and stores each line in the following Node. Then, once the file reaches its end, I run through the linked list and print all of the lines.
However, when I test it, it only prints blank lines, except for the last line in the file, which gets printed normally. In addition, despite the fact that all the strings are malloc'd before they are stored in the nodes, I get a "pointer being free was not allocated error." I've gone through this pretty extensively in gdb and can't seem to figure out what I'm doing wrong. Perhaps somebody else can help me out here? Here's the rest of my code:
int main(int argc, char **argv) {
if (argc>1) {
FILE *mfile = fopen(argv[1], "r");
if (mfile!=NULL) {
char c;
char *s = (char*) malloc(1);
s[0] = '\0';
LinkedList *lines = (LinkedList*) malloc(sizeof(LinkedList));
while ((c=fgetc(mfile))!=EOF) {
if (c=='\n') {
setNextLine(lines, s);
free(s);
s = (char*) malloc(1);
s[0] = '\0';
}
else s = append(s, c);
}
if (strlen(s)>0) {
setNextLine(lines, s);
free(s);
}
fclose(mfile);
printList(lines);
LLfree(lines);
} else perror("Invalid filepath specified");
} else perror("No input file specified");
return 0;
}
void setNextLine(LinkedList *lines, char *line) {
struct Node **root = &(lines->head);
while (*root!=NULL) root = &((*root)->child);
*root = (Node*) malloc(sizeof(Node));
(*root)->child = NULL;
(*root)->value = line;
}
char *append(char *s, char c) {
int nl = strlen(s)+2;
char *retval = (char*) malloc(nl);
strcpy(retval, s);
retval[nl-2] = c;
retval[nl-1] = '\0';
free(s);
return retval;
}
void printList(LinkedList *lines) {
Node *root = lines->head;
while (root!=NULL) {
char *s = (char*) root->value;
printf("%s \n", s);
root = root->child;
}
}
void LLfree(LinkedList *list) {
if (list->head!=NULL) NodeFree(list->head);
free(list);
return;
}
void NodeFree(Node *head) {
if (head->child!=NULL) NodeFree(head->child);
free(head->value);
free(head);
return;
}

Answer 1

It appears that there are several things that could be changed in the code.
Perhaps the one that is most likely to help would be that memory improperly freed.
Change:
setNextLine(lines, s);
free(s);
s = (char*) malloc(1);
to:
setNextLine(lines, s);
// free(s);
s = (char*) malloc(1);
The pointer 's' is still pointing to what was just assigned to the previous node's 'value'. Hence, calling 'free(s)' is actually freeing the node's 'value'.

Answer 2

Try doing this
void NodeFree(Node *head)
if (head->child!=NULL)
NodeFree(head->child);
free(head->value);
free(head->child);
free(head);
head->value = NULL;
head->child = NULL;
head = NULL;
return;
}

Answer 3

The setNextLine() function is appending the 's' poitner to the node value, and then that same pointer is getting freed after that call in the while loop.
That's why you'll get a double free fault when NodeFree() tries to free head->value.
And the fact that you get the last line could be just because the address pointed by 's' for the last line (which got freed like all the previous lines) is still unused although its not allocated to your pointer anymore.
You should make a copy of the line pointed by 's' in setNextLine() so you can work with the 's' pointer for the rest of lines.