Thread: linked list

Looks about right, Salem. Here's the code in the an acceptable form:

Code:

temp = list1;
while (temp->link != NULL)
{
	temp = temp->link;
}
temp->link = list2;

Note how the [code] tag make it readable? And don't forget the closing tag either.

Originally posted by daveS
I'm working on linked lists too, and came across a symbol I just cant understand. Its this: " -> " what does it mean?

Here is the peice of code its in, that I'm trying to understand.

Code:

new = (person*)malloc(sizeof(struct person));
new->next = head;
head = new

And also, can anyone translate that first line? I still don't under stand how the * sign can be after person.
Thanks for any help.

The first line allocates memory. malloc returns a void * so they cast it to a person *. The -> is used to dereference a pointer AND access a member. Another way to write it is:

*new.next = head;

Note: If you don't have a really good grasp on pointers I would suggest reading up on them before jumping into linked lists.

Originally posted by daveS

Code:

new = (person*)malloc(sizeof(struct person));
new->next = head;

Another way of saying what MrWizard said:

First:
malloc() returns a pointer (*) but it is of type void. You need it as a pointer to a type 'person' therefore (person *) is used to redefine the pointer to the proper type.

Second:
If you have defined:
person new;
new is a structure, you reference it's members as new.member

But you have new as a pointer to a structure person. You therefore need to use new->member to reference thru the pointer to the structure.

I always thought the . (dot) operator had prescedence over the * (derefernce) operator so wouldnt you need to do itl ike this?

Code:

(*person).member