Binary tree passing by reference maybe?

Question

first time commenter here. I'm trying to add char arrays to a Binary Tree in java. The problem is that when I try to make the char array into a node (prior to adding the node to the tree), the node is not a char array but a reference. Here is my making-things-a-node class:

public class TreeNode {

char[] data;
TreeNode left, right, parent;
public TreeNode(char[] data) {
    this.data = data;
    left = right = parent = null;
}
}

and here is the line I'm using to pass the char array to the node class:

TreeNode theNode = new TreeNode(dictionary);

where dictionary is a char array. And when I print theNode I get something like this: ([C@58623a98). I think the problem might be with the this.data but I'm not sure. Can anyone help me?

Update: when I print dictionary I get the char array I started with.

If you want to print the content of data, use `Arrays.toString(data);` — Alexis C., Mar 29 '14 at 19:13
Look into printing arrays and the inherited `Object#toString()` method. — Sotirios Delimanolis, Mar 29 '14 at 19:13
The problem is not the printing of the array. That works fine. The problem is making the array into a node. Also, how would I go about making a copy in this particular case? — jani, Mar 29 '14 at 19:24

score 2 · Accepted Answer · answered Mar 29 '14 at 19:17

What you describe is normal behaviour. The outcome of printing an object is determined by its .toString() method. If no .toString() method is declared, the default .toString() method inherited from Object class is used, which gives the output you received.

If you want the array to look nice when printed, you can define your own .toString() method, or use an utility method of Arrays.toString(yourArray).

score 0 · Answer 2 · edited May 23 '17 at 12:05

0

The observation is correct that each node has access to the same character array¹; if you want a copy of any object, make a copy of an object.

Either of the following are sufficient to make a copy of a primitive array, although clone has a bad reputation in general.

this.data = Arrays.copyOf(data, data.length);
this.data = data.clone();

Alternatively, you could make it the callers responsibility to pass in different objects.

// i.e. when omitting the above
new TreeNode(dictionary.clone());

More complex objects/object-graphs will require different handling. My recommendation is to simply try and reduce mutable objects which eliminates the need for making the appropriate copies.

¹ That is, objects are not copied/cloned/duplicated when they are passed in Java; rather, the value of the reference to the object is passed, and hence the same object is referred to.

edited May 23 '17 at 12:05

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answered Mar 29 '14 at 19:26

user2864740

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That did not work, but it seems like it should. I still have the same output as before. Perhaps my problem lies not where I thought it did. – jani Mar 29 '14 at 19:37
@jani Oh, I see. Your printing issue is *unrelated* to the above. – user2864740 Mar 29 '14 at 19:38
@jani See http://stackoverflow.com/questions/409784/simplest-way-to-print-an-array-in-java – user2864740 Mar 29 '14 at 19:41

Binary tree passing by reference maybe?

2 Answers2