Print linked list up to specific number in C - c

Question

Suppose I have a linked list of unknown length. I would like to have a function to print out whole list if the length is less than 10, and if its length more than 10, then display only first 10 nodes.
However, because I'm comparing the pointer and integer to print out first 10 nodes, I'm getting the following error:
ordered comparison between pointer and integer ('NodeT *' (aka 'struct node *') and 'int')
for (current = list; current < 30; current = current->next)
If I change it to have a count and loop while count is less than 10, I'm getting a segmentation fault.
How can I display first 10 nodes if list is length is more than 10?
I have the following function:
void *printNodes(NodeT *list) {
int length = countNodes(list); // finds length of a linked list
int count = 0;
NodeT *current;
if (list == NULL) {
printf("Empty list.\n");
}
if (length < 10) {
// Display all if number of nodes is less than 10
for (current = list; current != NULL; current = current->next) {
printf("%s\n", current->data);
}
} else {
// Display first 10 if number of nodes more than 10
for (current = list; current < 10; current = current->next) {
printf("%s\n", current->data);
// for (current = list; count < 10; current = current->next) {
// printf("%s\n", current->data);
// count++;
}
}
return 0;
}

Answer 1

For starters the return type void * of the function printNodes
void *printNodes(NodeT *list) {
does not make a sense.
You should declare the function at least like
void printNodes( const NodeT *list) {
As the number of potentially outputted nodes is known there is no need to determine how many nodes there are in the list.
If the condition of this if statement
if (list == NULL) {
printf("Empty list.\n");
}
evaluates to the logical true the function shall return.
The condition of the for loop where a pointer is compared with an integer
for (current = list; current < 10; current = current->next) {
does not make a sense.
The function can be declared and defined the following way.
void printNodes( const NodeT *list )
{
size_t n = 10;
if ( list == NULL )
{
printf("Empty list.\n");
}
else
{
for ( ; list != NULL && n--; list = list->next )
{
printf( "%s\n", list->data );
}
}
}
Another approach is to declare one more parameter that will specify how many nodes of the list you want to output. For example
void printNodes( const NodeT *list, size_t n )
{
if ( list == NULL )
{
printf("Empty list.\n");
}
else
{
for ( ; list != NULL && n--; list = list->next )
{
printf( "%s\n", list->data );
}
}
}

Answer 2

Consider adapting your main for loop to serve as a counter:
void printNodes(NodeT *list) {
if (list == NULL) {
printf("Empty list.\n");
return; // Note no more useless "pointer" return
}
int limit = 10; // Could easily be an argument
NodeT *current = list;
for (int i = 0; current && i < limit; ++i, current = current->next) {
printf("%s\n", current->data);
}
}
The key here is you need to test that both the pointer is valid and the counter hasn't tripped.
In your adaptation you walked off the end of the chain without paying attention to your current pointer.

Answer 3

You can try the else part in this way:
for (current = list; count< 10; count++, current = current->next)
///your code

Answer 4

You can use a for loop as mentioned by tadman or a while loop like this
int counter = 0;
NodeT *current = list;
while (counter < 10 && current != NULL)
{
/* Do your stuff */
current = current->next;
counter += 1;
}
Instead of finding the length of the linked list as you did, which might make the time complexity of your function linear, you just need to check if the counter is at most ten and if the current node is not null.