Thread: Using a linked list globally and File I/O

I would have thought that you do the loop through the linked list in printall, and printone is responsible for printing a particular node. Your model appears to imply that it's printing the "nth" element in the linked list, and presumably traverses the list every time.

"Pointer to incomplete type" would indicate that you don't have a "struct node" declared - perhaps it's called something else?

--
Mats

I saw that too, and I've changed the code. printall has the loop to go to the next node before handing "current" over to printone so that printtone can actually print the data in "current", the Nth node.

Another quick question though.

Code:

current->soldier = soldiers

works, while...

Code:

soldiers = current->soldier

doesn't. Why?

EDIT: Full code.

Code:

void printall(struct node** head)
{
	struct node* current = *head;
	
	printf("\n");
	
	while (current->next != NULL)
	{
		printone(&current);
		current = current->next;
	}
}

void printone(struct node** current)
{
	soldiers = current->soldier;
	printf("\n\n");
	printf("ASM: %ld", soldiers.aa);
	printf("Onomatepwnymo: %s", soldiers.firstname);
}

Originally Posted by Leftos

I saw that too, and I've changed the code. printall has the loop to go to the next node before handing "current" over to printone so that printtone can actually print the data in "current", the Nth node.

Another quick question though.

Code:

current->soldier = soldiers

works, while...

Code:

soldiers = current->soldier

doesn't. Why?

EDIT: Full code.

Code:

void printall(struct node** head)
{
	struct node* current = *head;
	
	printf("\n");
	
	while (current->next != NULL)
	{
		printone(&current);
		current = current->next;
	}
}
Are you sure this is correct - I expect this to not print the final node.

void printone(struct node** current)
{
	soldiers = current->soldier;
	printf("\n\n");
	printf("ASM: %ld", soldiers.aa);
	printf("Onomatepwnymo: %s", soldiers.firstname);
}

What is soldiers - is it a global variable?

See also red comment above.

--
Mats

Thanks, that worked. I hope everything goes well from here on now. And thanks to everyone who bothered to show up and reply.

Originally Posted by Leftos

So now I know that I need to do this...

Code:

void printall (struct node** head)

...in order for the printall function to have access to the list. However, if I call printone from within printall and I want printone to have access to the same list as well, how do I declare printone, and how do I pass on head?

No you don't.
You only need ** if a function is going to modify a *. So in essence, taking a struct node** head means it's going to reassign the head * pointer, which it shouldn't.

Okay, well, the double asterisk is only needed when the called function is going to modify the linked list.

Any ways to delete an element from a single linked list? I'm confused as to how I'll set the previous element's pointer to point to the next element from the one deleted.

I tried the following but I get segmentation fault.

Code:

void delrec(struct node** head, int acount)
{
	int i;
	struct node* current = *head;
	struct node* todel;
	
	for (i=0;i<=acount-1;i++)
	{
		current = current->next;
	}
	
	todel = current->next;
	current->next = current->next->next;
	free(todel);
}

robwhit, I based my code on the tutorials I found in this e-book. Pages 12 to 14 are what I based the double asterisk on. If they're wrong or I'm still wrong, please do tell.

Could this work?

Code:

void delrec(struct node** head, int acount)
{
	int i;
	struct node* current = *head;
	
	for (i=0;i<=acount-1;i++)
	{
		current = current->next;
	}
	
	current->soldier = current->next->soldier;
	current->next = current->next->next;
	free(current->next);
}

Nope. Segmentation fault. I obviously can't access current->next->soldier.

EDIT: SOLVED! Took a look here, and changed it only slightly for it to work with my code.

Code:

void delrec(struct node** head, int acount)
{
	int i;
	struct node* current = *head;
	struct node* previous = NULL;
	
	for (i=1;i<=acount-1;i++)
	{
		previous = current;
		current = current->next;
	}
	
	if (previous == NULL) *head = current->next;
	else previous->next = current->next;
	free(current);
}

The double asterisk is only needed if you want to be able to modify the head variable that you have in the calling code. It will still point to the same linked list.

You could do it that way, but gaining the extra level of indirection just to dereference it again is extraneous.

Singly linked lists don't have a previous member. edit: haha never mind that.

Suppose that you delete more than one node at a time. How many times would you have called malloc? How many times would you have called free? edit2: nevermind to that too...