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Im having trouble with a constructor for a linked list. it takes a string and is supposed to create a node for every character.

i get a nullpointerexception everytime i try to print out the list. Does that mean that not even the first node is being created?

below is my node class and the list constructor.

class CharNode { 

   private char letter; 
   private CharNode next; 
   public CharNode(char ch, CharNode link)
    { 
        ch = letter;
        link = next;
    }

   public void setCharacter(char ch) 
    { 
        ch = letter;
    }

   public char getCharacter() 
    { 
    return letter;
    }
   public void setNext(CharNode next) 
    { 
    this.next = next;
    }

   public CharNode getNext() 
    {
    return next;
    }    
}

and this is my constructor

   // constructor from a String 
   public CharList(String s) { 

    CharNode newNode = head;

    for(int i = 0; i <s.length(); i++)
    {
        newNode = new CharNode(s.charAt(i), null);
        newNode.setNext(newNode);
    }

    }

am i constructing it correctly?

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2 Answers 2

Reset to default
2

First of all, your attributions are switched around!

ch = letter;
link = next;

should be

letter = ch;
next = link;

Same thing in your setter.

When you have a method in Java of the form:

public void setSomething(String argument){
    this.classMember = argument;
}

Is usually what you want. You must assign your argument to your class member, not the other way around.

Also, when you're invoking your constructor, you have:

newNode = new CharNode(s.charAt(i), null);
newNode.setNext(newNode);

This make it so that your "link" is always pointing to itself! Think about what you need to do to make the previous Node point to the node you've just created (perhaps save it somehow?)!

Was I clear enough? Let me know if I can explain further.

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5 Comments

thanks, i thought that the way i had it looked a bit off but i wasnt sure. it makes more sense to me to have letter = ch
No prob :) hope it gets you on the right track! You can accept the answer if it solves your problem. Have fun!
would i have to create a new CharNode for each character?
Yes, I believe you're doing that part right. You just need to create a variable outside of your cycle where you'll keep the lest Node you've created, so you can always update it with the newly created node (forming a linked list). You can check this out: en.wikipedia.org/wiki/Linked_list if in doubt. Good luck!
a variable like a count? I'm sorry, this part just really throws me off
0

As pcalcao said, = assigns the value on the right to the variable on the left. You'll need to change ch = letter; to letter = ch; and link = next; to next = link;

Now, the line CharNode newNode = head; doesn't mean anything unless you've specified what head is prior to the code you've given, but it doesn't look like it. Remember that when creating a linked list, you don't start with anything, so even "special" nodes like head must be created (instantiated, if you prefer). The idea is to create the first node (the head), and then assign head to that first node. For every node after the first one, this step isn't required, as you simply are adding to the end of the list.

Finally, during construction, you'll need a reference to both a new node (like you have now), and the previous node, in order to do proper appending. Your code now is simply setting the next node in the list to the current node, which means that once you create a new newNode, you will lose the reference to the previous newNode.

A good way to approach linked lists when one is first getting started is to draw out what you want to do step by step, then try and translate that into code. Hopefully this is helpful.

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2 Comments

is each node going to have a different name? like would the first be newNode and the second newNode1?
No, otherwise you would have trouble making arbitrary long lists. If I have two references to nodes, say curNode and prevNode, I can loop, doing something like prevNode.next = curNode;. After this, I can move prevNode to reference curNode, and now I am free to create a new node and assign curNode to it. As long as I have the first node in the list (head), I can access the whole list.

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