循环单链列表中任何选定索引的返回指针

Question

I designed this code so that I can get a pointer of any position that user wants in a circular singly linked list, I am using cout to return pointer, I want such a mechanism that I can use it with my other function instead of re writing the whole code again, for that I need to do something with return type which is void right now

Here is the function ..

void pointer_to_node(int index){
    Node*temptr;
    temptr = new Node;
    temptr = firstptr;

    Node*temptr2;
    temptr2 = new Node;
    temptr2 = NULL;
    int count = 1;

    while (temptr!=temptr2){
        if(count==index){
            cout << "Required Pointer is : ";
            cout<< temptr;}

        count++;
        temptr2=firstptr;
        temptr=temptr->nextptr;
    }

    if (index>size_of_list())
    {
        temptr=NULL;
        cout<< "Can't You think in bounds. Take your NULL Pointer ";
        cout << temptr;
        delete temptr;
        delete temptr2;
    }
}

Answer 1

You just need to return a Node * .

However, while you're doing that, you also really need to take out these: temptr = new Node; lines, as well as the delete s, as you're leaking memory there. You just immediately discard those new nodes by reassigning the pointers. The delete s at the end will delete the wrong nodes entirely, and aren't called in all cases anyway.

And if you pass an index of 0, your loop might take a very long time indeed.

I assume you have good reason for wanting to return NULL if you loop around the list.

Something like the following should suffice:

Node *pointer_to_node(int index)
{
    Node *temp = firstptr;
    while(index-- != 0) {
        temp = temp->nextPtr;
        if(temp == firstptr) return NULL;
    }
    return temp;
}