Function with generic parameter java

Question

I have a tree of objects T, declared as

 public class Tree<T> {

    private Node<T> root;

    public Tree(T rootData) {
        root = new Node<T>();
        root.data = rootData;
        root.children = new ArrayList<Node<T>>();
    }

    public static class Node<T> {
        private T data;
        private Node<T> parent;
        private List<Node<T>> children;
    }
}

and want to add a function keepBranch, which reduces the tree to one of its branch.

But I need keepBranch to take an object T as a parameter, to select the branch.
Something like:

public void keepBranch(<T> param) {

        for (Node<T> node : this.root.children) {
            if (param.equals(node.data)) {
                this.root = node;
            }
        }
}

Is there a way to do this? Or am I doing it wrong?

<T> is used to define a type variable in a certain context. But once defined, you can use T as you would use any other class name. (Syntactically, that is. You can't do everything you could do with a concrete class name.) — biziclop
– biziclop, Commented Feb 9, 2017 at 14:21
You already have an example for the correct usage in your constructor public Tree(T rootData), so I wonder why you've thought that you need to use <T>. — Tom
– Tom, Commented Feb 9, 2017 at 14:26
@Tom Took the tree code from another thread, as I'm beginning with generics and trying to understand it. But yeah you're right I should have noticed! — M. Garrigues
– M. Garrigues, Commented Feb 9, 2017 at 14:28
data in our code should be unique for whole tree. if it isn't so you can get wrong branch by data comparison. — Vlad Bochenin
– Vlad Bochenin, Commented Feb 9, 2017 at 14:29
@VladBochenin You're right, I will be carefull when using it, thanks! — M. Garrigues
– M. Garrigues, Commented Feb 9, 2017 at 14:37

f1sh · Accepted Answer · 2017-02-09 14:16:38Z

4

Change the parameter type to T. But your comparison is wrong, you compare a value to a node. Change is so that the node's value is being compared to param:

public void keepBranch(T param) {
    for (Node<T> node : this.root.children) {
        if (param.equals(node.data)) {
            this.root = node;
        }
    }
}

answered Feb 9, 2017 at 14:16

f1sh

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2 Comments

Vlad Bochenin Over a year ago

OP maybe wants to compare by reference. For example some nodes could be equals but placed on different places in tree

biziclop Over a year ago

@VladBochenin It isn't the reference comparison that is the biggest problem, it's comparing a Node instance (node) to something that is definitely not a Node instance (param).

Mickael · Accepted Answer · 2017-02-09 14:21:27Z

2

It should be T param instead of <T> param :

public void keepBranch(T param) {

        for (Node<T> node : this.root.children) {
            if (param.equals(node.data)) {
                this.root = node;
            }
        }
}

EDIT

As said in comments, it should be param.equals(node.data) instead of node == param

Documentation : Lesson: Generics (Updated)

edited Feb 9, 2017 at 14:21

answered Feb 9, 2017 at 14:15

Mickael

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4 Comments

jack jay Over a year ago

node == param is comparing Node<T> == T. is it correct?

f1sh Over a year ago

@jackjay no, it's not. It has to be compared to node.data.

M. Garrigues Over a year ago

Well I just took this comparison for the example, I can edit it if it bothers you though.

M. Garrigues Over a year ago

@jackjay No, I just thought of passing a parameter of type T, to compare it to the node data. Should I pass a Node<T> type instead?

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