While loop with pointers does not work

Question

I have a problem, I am trying to create a list that deletes a highest value holding number, or all numbers with the same value if the value is highest in the list. Thank you for any kind of tips.

// n,n1,head,next - are pointers
int j = 0; //this number helps to put pointer forward by one place
while(n!=0){//should go through every digit of the list
    if(head == 0){
        cout << "list is empty" << endl;
    }
    else{
        n = head;
        n1=0; // n1 and n are pointers
        while(n!=0){
            if(n->sk == maxx){//searches for maximum digit in the list
                break;
            }
            else{
                n1=n;
                n=n->next;
            }
        }
        if(head == n){
            head = head->next;
        }
        else{
            n1->next = n->next;
        }
        delete n; // deletes the pointer holding the highest value
    }
    n = head; //problem is here or somewhere below
    j++;
    for(int i=0; i<j;i++){ // this loop should make the pointer point to the first
        n = n->next;       // number, then the second and so on until the end of list
    }                      // and all the numbers inside the list with the value that
}                      // equals "maxx" should be deleted

Poriferous · Accepted Answer · 2014-12-05 22:30:15Z

2

You should dereference the pointers. Right now, you're pointing to their addresses. See if that helps resolve your problem.

answered Dec 5, 2014 at 22:30

Poriferous

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Raydel Miranda · Accepted Answer · 2014-12-05 23:04:05Z

0

Ok, the problem (the most of it) is the code:

   while(n!=0){
        if(n->sk == maxx){
            break;
        }
        else{
            n1=n;
            n=n->next;
        }
    }

If you find the maxx value you should delete that node and continue to searching, don't break. This way you don't need so much code for this task.

while (n != 0){
    if (n->sk == maxx){
        node *prev = n->prev; // The previous node.
        node *tmp = n;        // this assume you have a class node.
                              // temporaly holds the pointer to n.
        prev->next = n->next; // Connect the previous node with the following one.
        n = n->next;          // advance n to the next node in the list.
        delete tmp;           // delete the node.
    }
}

edited Dec 5, 2014 at 23:04

answered Dec 5, 2014 at 22:41

Raydel Miranda

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4 Comments

Benas Ročas Over a year ago

Would your answer still be the same if the list is two-sided?

Raydel Miranda Over a year ago

By two-sided you mean circular?

Benas Ročas Over a year ago

i mean double linked list if my english is right where you can move forward and backward with pointers

Raydel Miranda Over a year ago

Yes my answer still working for double linked lists. You can still move over the entire list from the head to the end. the main issue with my answer is that don't take in to account that the list has to remain conected, I will fix that right now!

tnull · Accepted Answer · 2014-12-05 22:46:50Z

0

If I understand correctly what you want, you can just iterate over your list and save the pointer for deletion:

it = head;
pos = nullptr;
while (it != nullptr) {
    if(it -> sk == maxx) {
        pos = it; // save ptr
        it = it -> next;
        delete pos; // delete saved ptr
        pos = nullptr;
    }
}

answered Dec 5, 2014 at 22:46

tnull

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