pointer to a pointer

[complete]

I need to brush up on pointers and I think a good way is to start with this question. If you want to change a pointer in a function, you need to pass a pointer to a pointer. WHy is this?




consider a linked list with the node like this:

typedef struct elementT{
int data;
struct element *next;
}


now, consider this function to insert a new header:


int BadInsert(element *head)
{
element *newElem;
newElem = (element *) malloc(sizeof(element));

if (!newElem)
return 0;

newElem->next = head;

head = newElem;

return 1;
}

This incorrectly updates local copy of head.
Calling code retains the old value for the first tlement
pointer, so it nowpoints at the second element of the list.

The correct way to update the head pointer in C is to pass a pointer
to the head pointer, allowing you to modify the calling function's pointer to the first element,
as shown here:

int Insert(element **head)
{
element *newElem;
newElem = (element *) malloc(sizeof(element));

if (!newElem)
return 0;

newElem->next = *head;

head* = newElem;

return 1;
}


My question is this. How would I call this Insert function?

[DJMaze]

Pass by reference:

Insert(&heads);

Problem is that you don't append but prepend instead. Therefore the pointer to "heads" is not altered.
A different way is:

element * prepend(element *head)
{
    element *newElem = (element *) malloc(sizeof(element));
    if (!newElem) return head;

    newElem->next = head;

    return newElem;
}

element *elements = (element *) malloc(sizeof(element));
elements = prepend(elements);

[landonmkelsey]

final exam...make sure you have it mastered!!!

I got this on an interview once!

write a little program method to switch the contents of two pointers

#include <iostream>
using namespace std;

void switchx(code goes here)
{
code goes here
}
int main()
{

int* p1 = new int();
*p1 = -100;
int* p2 = new int();
*p2 = 100;
cout<<"before "<<*p1<<" ,"<<*p2<<endl;
switchx(&p1, &p2);
cout<<"after "<<*p1<<" ,"<<*p2<<endl;
}

answer on request!