Question

I was writing this code for adding element at the end of linked list:

struct node{
    int info;
    struct node* link;
};

void append ( struct node **q, int num )  
{

struct node *temp, *r ;

if ( *q == NULL )       // if the list is empty, create first node
{
    temp = (struct node*) malloc ( sizeof ( struct node ) ) ;
    temp -> info = num ;
    temp -> link = NULL ;
    *q = temp ;        
}
else{
    temp = *q ;         

    /* go to last node */
    while ( temp -> link != NULL )
        temp = temp -> link ;

    /* add node at the end */
    r = (struct node *)malloc ( sizeof ( struct node ) ) ;
    r -> info = num ;
    r -> link = NULL ;
    temp -> link = r ;
}
}

and I call append function like this:
append(&list, 10); where list is the pointer to the linked list

This code works, but if I use single pointer in append function(using *q instead of **q) and make changes accordingly (as done below and also when I call it), it doesn’t work. What is wrong with the code below?:

void append ( struct node *q, int num )  
{

struct node *temp, *r ;

if ( q == NULL )       // if the list is empty, create first node
{
    temp = (struct node*) malloc ( sizeof ( struct node ) ) ;
    temp -> info = num ;
    temp -> link = NULL ;
    q = temp ;        
}
else{
    temp = q ;         

    /* go to last node */
    while ( temp -> link != NULL )
        temp = temp -> link ;

    /* add node at the end */
    r = (struct node *)malloc ( sizeof ( struct node ) ) ;
    r -> info = num ;
    r -> link = NULL ;
    temp -> link = r ;
}
}

Answer 1

Because in the second example, q is a copy of the pointer passed in by the caller. The caller’s original pointer never gets modified.