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Use method reference with parameter

I just started learning Java streams and faced a problem. Please take a look at a the following example. This is part of a Node class:

private Map<String, Node> nodes;  public Optional<Node> child(String name) {     return Optional.<Node>ofNullable(nodes.get(name)); }  private void findChildren(String name, List<Node> result) {     child(name).ifPresent(result::add);     nodes.values().stream() //          .map(Node::findChildren(name, result)) //          .forEach(Node::findChildren(name, result))             .forEach(node -> node.findChildren(name, result)); } 

My intent was to call #findChildren with the name and result parameters on each node in the stream. I tried to use the method references Node::findChildren with no luck. I'd appreciate solutions other the the one with -> operator.

Is it somehow possible to use the method reference together with a parameter? I like the idea of streams and I just want to make the code more readable.

Actually, I think there is a similar question Method references with a parameter which I read but cannot figure out how to use the bind2 method in my code. Is it the only solution?

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matepal297 Avatar asked Apr 23 '15 22:04

matepal297


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When can I use method reference?

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1 Answers

You can’t use method references for this purpose. You have to resort to lambda expressions. The reason why the bind2 method of the linked question doesn’t work is that you are actually trying to bind two parameters to convert a three-arg function into a one-arg function. There is no similarly simple solution as there is no standard functional interface for three-arg consumers.

It would have to look like

interface ThreeConsumer<T, U, V> {     void accept(T t, U u, V v); } public static <T, U, V> Consumer<T> bind2and3(                         ThreeConsumer<? super T, U, V> c, U arg2, V arg3) {     return (arg1) -> c.accept(arg1, arg2, arg3); } 

Then .forEach(bind2and3(Node::findChildren, name, result)); could work. But is this really simpler than .forEach(node -> node.findChildren(name, result));?

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Holger Avatar answered Oct 13 '22 12:10

Holger