C: getchar() destroys pointer - c

Question

typedef struct LinkedList LinkedList;
struct LinkedList {
LinkedList* next;
char* head;
char current;
};
LinkedList makeList()
{
char* headPointer = calloc(60, sizeof(char));
LinkedList temp = { 0xCCCCCCCC, headPointer, 0 };
return temp;
}
int addToList(LinkedList* lstPointer, char toAdd) {
if (lstPointer->head == NULL || lstPointer->head == 0xCCCCCCCC)
return -1;
if (lstPointer->current + 1 < 60) { /* enough space in the list to add */
*(lstPointer-> head + lstPointer -> current) = toAdd;
lstPointer->current = lstPointer->current + 1;
}
else /* not enough space, will create new node in the list */
{
if (lstPointer->next == 0xCCCCCCCC) {
LinkedList nextNode = makeList();
lstPointer->next = &nextNode;
}
return addToList(lstPointer->next, toAdd);
}
/*Added succsessfully*/
return 0;
}
int main(){
char chr;
LinkedList lst = makeList();
while ((chr = getchar()) != EOF) {
if (addToList(&lst, chr) == -1)
return -1;
}
return 0;
}
i am trying to use linked list but after i fill the first, i create a new one and able to add an item to it. on the second item the next list pointer get destroyed by getchar(). i have no idea why or how is it related.

Answer 1

In makelist you need to allocate a new list, but then instead of returning it, you copy it into a local variable, leaking the memory that you just allocated. Instead, return a pointer:
LinkedList *makeList() // Note *
{
LinkedList *temp = calloc(1, sizeof(LinkedList));
temp->head = calloc(60, sizeof(char));
temp->next = 0;
temp->current = toAdd;
return temp; // Note temp is a pointer
}
In addToList you don't need the nextNode variable:
lstPointer->next = makelist();