C Linked lists with global variables - c

Question

Hi I'm new to C and I'm struggling to get a linked list working I'm having to use global variables because I can't change the parameters of the function that is being called. I declare the struct and two global pointers to keep track of the root address and last address like this.
struct node {
char* pointer;
struct node *next;
};
struct node** rootNode;
struct node** lastNode;
Then inside the function I malloc a new struct and try to setup the first node in the list. To try and save the location of this node I think I'm assigning the global pointer lastNode to the pointer of root.
struct node *root = malloc(sizeof(struct node));
...
root->pointer=ptr;
root->next = 0;
rootNode = &root;
lastNode = &root;
Then I try to add aditional nodes by mallocing another node and then linking the previous node using the address stored in my lastNode pointer.
struct node *newNode = malloc(sizeof(struct node));
newNode->pointer=ptr ;
(*lastNode)->next = newNode;
newNode->next = 0;
lastNode = &newNode;
However this doesn't really seem to work. When I run the following the program matches the first two items in the list but then returns null for all nodes after that. I've been stuck on this for 2 days now and any help would really be appreciated :)
struct node* test;
test = (*rootNode);
enter code here
while (test) {
if (test->pointer == ptr) {
printf("MATCH!!");
notFound = 0;
break;
}
else {
test = test->next;
}
}
EDIT A couple of people have asked me to supply some more code. This is the function in which I would like to create the linked list. It's called multiple times while my program is running and I'm trying to add a new node to my linked list every time it gets called. I've also included the global variables at the top.
struct node** rootNode;
struct node** lastNode;
int firstRun = 1;
struct node {
char* pointer;
struct node *next;
};
void malloc(size_t sz) {
size_t maxSize = (size_t)-1;
char * payloadPtr = NULL;
if (sz > maxSize - sizeof(struct node)+sizeof(int)) {
return ptr;
}
if (firstRun) {
struct node *root = malloc(sizeof(struct node));
ptr = malloc(sizeof(size_t)+sz);
if (ptr == NULL) {
return ptr;
}
memcpy(ptr, &sz, sizeof(int));
payloadPtr = ptr+1;
root->pointer=payloadPtr;
root->next = 0;
rootNode = &root;
lastNode = &root;
firstRun = 0;
}
else {
struct node *newNode = malloc(sizeof(struct node));
ptr = malloc(sizeof(size_t)+sz);
if (ptr == NULL) {
return ptr;
}
memcpy(ptr, &sz, sizeof(int));
payloadPtr =ptr+1;
newNode->pointer= payloadPtr;
(*lastNode)->next = newNode;
newNode->next = 0;
lastNode = &newNode;
}
return payloadPtr;
}

Answer 1

There's a little clarification about changing the parameters into a function. In C you can only pass by pointers on function calls so if you want to modify the value of a struct node - you pass it like struct node * in the function. Similarly if you want to change the value of a struct node * in the function (as in allocate it or delete it) you may want to pass it like struct node **.
Your function prototype will probably have to look like this:
void addNode(struct node** root, char* value)
But you can also make that root node a local and also just a struct node * instead of struct node **. And then simply call it like:
addNode(&root, value);