CODEWITHSUNDEEP
cpointers
user3818219
user3818219

Reputation: 43

head pointer to pointer

I found this linked list routine (from Unix System Programming by Haviland) to add an item to a singly linked list.

additem(item **head, item *newitem)
{
  newitem->next = *head;
  *head = newitem;
}

My difficulty is in trying to understand the item **head part. What does the **head really mean? Why not define the routine with just item *head as follows:

additem(item *head, item *newitem)
{
  newitem->next = head;
  head = newitem;
}

Upvotes: 2

Views: 112

Answers (5)

Nikole
Nikole

Reputation: 4749

http://cslibrary.stanford.edu/102/PointersAndMemory.pdf may be helpful for you to understand.

Upvotes: 1

Long_GM
Long_GM

Reputation: 191

Just understand simply that: if you want another function to change the value of a variable. You have to pass its address since C use "call by value" mechanism.

void change_value(type *a) 
{
    *a = ...;
}

In this case, you want to change the value of a pointer to head type (head*), so its addr has the type of head**

type * a equal to (head*) *a or head **a

Upvotes: 0

Havenard
Havenard

Reputation: 27894

It is a pointer to a pointer.

In C you cannot change variables passed by value, as parameters are handled as local variables. You have to pass their pointers instead, and then modify the value of *variable.

It also applies when the variable you want to change is already a pointer. In this case you have to pass the pointer to that pointer, so when you change this pointer, this change propagates to the caller.

*head is the value head is pointing to. while head is of type item **, *head is of type item *.

When you do this:

additem(item *head, item *newitem)
{
  newitem->next = head;
  head = newitem;
}

Modifying the value of head does not affect it's value to the caller. To all purposes it is meaningless to change head in this context.

Upvotes: 3

Yu Hao
Yu Hao

Reputation: 122443

You have the answer of your first question in your title: item **head means head is a pointer to a pointer.

Why your second code snippet doesn't work? because 

 head = newitem;

assigns the value to the local variable head, which is not what the code is intended to do.

Upvotes: 1

JohnH
JohnH

Reputation: 2721

If you just passed the pointer *head to the child routine and change its value and try to return it, you will discover you cannot because it is a value passed to the routine which only exists on the routine's stack as long as the routine is in scope. However, you can pass a pointer to that pointer (**head, which will be local on the stack as well), and change the value of what it points to, and then the value you want to be changed in the scope of the calling routine is actually changed.

This is the 'C way' of achieving call-by-reference even though C only officially supports call-by-value — it uses a call-by-value with a value that is a reference to the thing you want to change, thus allowing call-by-reference.

Upvotes: 0

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