Question

Alright guys, so I’ve been working on a linked list program and I’ve run into a problem. I’m trying to place a piece of data into an already sorted linked list in a sorted fashion. Right now, I’m comparing the pointer data the user gives (Data * newData) to the head (LinkNode * current = head) and I know that’s where my program is failing. I know I have to compare actual values, not memory addresses, but I’m not sure how I would go about doing it. Anybody have any suggestions at all or ideas?

Thanks.

void addSorted(Data * newData)
{
    if(head == NULL)
    {
        head = new LinkNode(newData);
        return;
    }
    LinkNode * current = head;
    LinkNode * previous = NULL;
    while(current != NULL)
    {
        if(newData->compareTo(current->data) == -1)
        {
            LinkNode * newNode = new LinkNode(newData);
            newNode->next = current;
            if(previous == NULL)
            {
                current->next = newNode;
            }
            else
            {
                newNode->next = previous->next;
                previous->next = newNode;
            }
            return;
        }
        previous = current;
        current = current->next;
    }
    previous->next = new LinkNode(newData);
}

Answer 1

Simply

    if(*newData < (*current->data))     

assuming that operator< is overloaded for the Data type

The minimal idiomatic way to provide std::less<> compliant weak total ordering (I.o.w. implement operator<):

struct Data
{
    int id;
    std::string other;

    // details omitted

    bool operator<(const Data& b) const
    {
         if (id < rhs.id)
             return true;
         if (id > ths.id)
             return false;
         return (other < b.other);
    }
};

If you have an large/complicated structure and all the members participating have comparison defined for their types, you can do this neat trick with TR1, Boost or C++11 Tuples:

    #include <tuple>

    // ...

        bool operator<(const Data& b) const
        {             
             return std::tie(id,other)< std::tie(b.id,b.other);
        }