在双链表C编程中使用反向查看/搜索功能的麻烦

Question

hey guys this is my first time doing double linked list so I'm not very sure what I'm doing here ,need some help to check the code,thanks, here is what I have done with comments included. The functions I have done here are print,print reverse,count elements in the linked list,and search function to determine whether this node exist

 void printListfow() //print the list in forward manner
{
    CLR;
    struct node *tmpval; //declare a temp storage
    if(start==NULL) //if 1st node = null,then nth is inside,nth to print
    {
         printf("List is empty\n");
         return; 
    }
     tmpval=start; //assign the head/start to temp storage to retrieve data in 1st node
     printf("List of customer details: \n");
     while(tmpval!=NULL) //keep doing till it is NULL/the end
     {
         printf("%d ", tmpval->detail); //print the 'detail' which is in the node temp is pointing at
         tmpval=tmpval->next; //assign next node to the temp storage so that it can be printed again
     }

}


void printListrev() //print in reverse manner
{
      CLR;
      struct node *tmpval; //temp storage
      if(start==NULL) //
      {
          printf("List is empty\n");
          return;
      }
      tmpval=start; //assign start to tmpval to retrieve value
      printf("List of customer details: \n");
      tmpval=tmpval->prev //move backward and assign the data to tmpval
      printf("%d",tmpval->detail) //print it

}

void count() //count total number of records
{   struct node *x;
    x=start; //assign value of start to temp storage
    int ctr=0; //initialize counter
    while(x!=NULL) 
  {
    x=x->next; //keep going to next node and then increase the counter
    ctr++;
  }
  printf("Number of customer records are %d\n",ctr);
}



int getNode(node *tmp ,int cust) //when user wants to delete a customer ID and its details, this will search through the list,then if found,pass the value to another function for deletion
{
    tmp=tmp->cust; 
    while(tmp!=NULL)
    {
        if(tmp->detail == cust) //check if detail[ID stored] is same as requested[cust]
        {
            return 1;
        }tmp=tmp->next; //if not same,then move to next one
    }return 0;

}

thanks!

Answer 1

In context to printListrev() :

Unless this is a circular doubly linked list, in which case last element is preceded by the first, start would have previous element NULL. So, there is no point in accessing the previous field of start , as you do here:

tmpval=start;
...
tmpval=tmpval->prev;

You can keep another pointer to end of the list for this purpose.

Other alternatives include:

recursive function:

void printrev(struct node *s)
{
    if (s == NULL)
    return;
    printrev(s->next);
    printf("%d ", s->detail);
}

iterative function:

void printrev()
{
    struct node *end;
    for (end = start; end->next != NULL; end = end->next)
    ;
    for (; end != NULL; end = end->prev)
    printf("%d ", end->detail);
}

Your getNode is of limited use. Suppose, if you want to delete element, your getnode would only return whether, element is present or not. Say it is present, your deleteNode function would still have to iterate to the appropriate element in the list before deleting it.

This could be solved by getNode returning pointer to the node:

node *getNode(int x)
{
    node *t;
    for (t = start; t != NULL; t = t->next)
        if (t->detail == x)
                return t;
    return t;
}

Now, you can code delete as follows:

void delNode(node *n)
{
    n->prev->next = n->next;
    n->next->prev = n->prev;
    free(n);
}

And call as follows:

node *n;

if ((n = getNode(x)) != NULL)
    delNode(n);

I've assumed that you struct is:

struct node {
    int detail;
    struct node *next;
    struct node *right;
};

typedef struct node * node;

Answer 2

• In printListrev() you are printing only one node not going reverse.
• One major problem you are doing is in getNode() ,you are changing the address of a local pointer but the original pointer still points where it previously.
• If so then how you are going to delete that node ,as you can't know the node address after this function returns.
• Are you going to call the getNode() for all the node ,if so that is not a good if you have many nodes.