Thread: Converting a program. From Vector to List

Hi,


I have a program, which uses a vector to store some pointers.

The program works fine.

I'm trying to convert it to use a doubly linked list now though.

At the top of the program I changed:

Code:

vector<Diary *>vectorname;

to

Code:

#include <list>

list<Diary *>vectorname;

list<Diary*>::iterator i = vectorname.begin();

Later on in the program I access an element of the (former) vector using:

Code:

if(vectorname[a]->getName() == leadToDelete)

And use:

Code:

if(advance(i,a)->getName() == leadToDelete)
{
      //do something
}

instead.

BUt when I try it fails to compile, saying:

error C2227: left of '->getName' must point to class/struct/union/generic type

I was hoping this would be a simple way to do the conversion, but I have made a mistake in the code above it seems.

Could anyone show me how I do this?

Many thanks for any help!

OK, I decided to start the program from the beginning again.

Using:

Code:

List<Base *>listname;

I am trying to delete a specific node from the list. Here is the case inwhich it happens:

Code:

case 'b':

	{
		//---------------------
		//Delete a 'Base'
		//---------------------
		cout << "You chose to delete" << endl;

		if(listname.size() == 0)
		{
			cout << "There are no nodes currently stored" << endl;
		}

		else
		{
			string NodeToDelete;
			cout << "Enter the datamember 'name' of the node to delete" << endl;
			getline (cin, NodeToDelete);

			for(unsigned int a = 0; a<listname.size(); ++a)
			{


					if(listname[a]->getName() == NodeToDelete)
					     {
					           cout <<"Found a match " << endl;

					          listname.remove(NodeToDelete);

					         cout << "Match deleted" << endl;
					     }
			}
		}
		break;
	}

I'm having trouble with the two lines in bold.

The first worked in the vector, but I know you can't access list elements like that.

The second line in bold is trying to remove the matching name using list::remove() but it say's:

'leadtodelete' : undeclared identifier

Code:

List<Base *>listname;
...
...
string NodeToDelete;
listname.remove(NodeToDelete);

These lines look incorrect. If listname is a list of Base pointers then remove must take in a Base pointer. You are telling it to remove a string yet you have not added strings to the list. If you meant to deref the iter and then grab the string from the object this is not exactly what you posted.

This may be more of what you were looking for. There aren't all that many differences between how you use or interface with the STL containers but there are significant differences in what they do and how they store their data. Once you get used to these differences you will be able to switch from list to vector to map to any other container with little or no trouble.

Code:

typedef std::list<Base *> BaseList;
typedef BaseList::iterator BaseListIter;
BaseList listname;
...
...
bool Foo::Remove(const std::string &NodeToDelete)
{
   bool result = false;
   BaseListIter iter(listname.begin());
   BaseListIter end(listname.end());
   
   while (iter != end)
   {
      if ( (*iter)->getName() == NodeToDelete )
      {
            //Cleanup the memory allocated by the object
            delete *iter;
          
            //Remove the object pointer from the list
            listname.erase(iter);
            result = true;
            break;
       }
       ++iter;
    }
    return result;
}

list::remove() will remove a unique element matching the type you pass in. However in your case your list contains Base pointers and not strings. Because remove() has no way of knowing that the Base object has a function that returns a string identifier and therefore will not use it as the determining factor for removal you must do some footwork to get it to do what you want. The code above, if it is correct as is, should iterate the list from start to finish and will check the name of the object via getName() which I'm assuming is in the Base object. If the name matches the name that is passed into the function the memory pointed to in the list will be cleaned up and list::erase() will erase the element which will remove it from the list. list::remove() will look for a Base pointer matching the one you specify b/c that is what you put into the list. If your function removed from the list based on Base pointer addresses then your removal would work. Your removal is using a different criteria or custom criteria for removal and thus list::remove() cannot be used.

'leadtodelete' : undeclared identifier

That has to be a typo since I do not see 'leadtodelete' anywhere in the code.


If you need random read/write access to the container list is probably not the best choice. Vector allows for random access and acts similar to an array. If you need to key off items by a unique key then std::map is a good choice. std::set also does this but you cannot alter the value at the key in std::set. Both set and map do not allow you to change the key.

Originally Posted by Swerve

I am trying to delete a specific node from the list. Here is the case inwhich it happens:

Wanting to perform deletions from the middle of a list as well as accessing items by index is almost certainly a logical error.

To me it indicates that a more appropriate container is a map. You can then access any item using a known key value, and potentially delete it, without affecting the ability to access any other item by its key value. The difference is that the key value is no longer its array index, but is instead some kind of cookie value..
I'd say it's time to learn about more containers.