Casting ArrayList<SuperClass> to ArrayList<SubClass>

Question

I know it sounds dumb, but I really need it. I'm 100% positive about only allowing SubClass into the arraylist. No other subclass of SuperClass is in that arraylist.

How can I do this?

    Map<String, List<Nodo>> map = ((Cast to ArrayList<NodoMoore> here)nodos).stream().
collect(Collectors.groupingBy(Nodo::getSalida));

My class implements an interface that makes me use Nodo instead of NodoMoore, but the getSalida method is not a Nodo method, only a NodoMoore method.

Can you show us a few skeleton class definitions that would help us replicate what you're seeing? Also, which class is implementing the interface? — Makoto
– Makoto, Commented Feb 22, 2016 at 7:19
Can you map first, eg map a Nodo -> a NodoMoore with a cast? — matt
– matt, Commented Feb 22, 2016 at 7:21

Eran · Accepted Answer · 2016-02-22 07:20:58Z

8

If you are sure that nodos contains only instances of NodoMoore, you can map the Nodo instances to NodoMoore instances with a cast :

Map<String, List<NodoMoore>> map = 
    nodos.stream()
         .map(n->(NodoMoore)n)
         .collect(Collectors.groupingBy(NodoMoore::getSalida));

I assumed that NodoMoore is a sub-class of Nodo.

answered Feb 22, 2016 at 7:20

Eran

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5 Comments

AFP_555 Over a year ago

Hello, DaoWen's answer has the output I want but this is also a very useful answer.

andred Over a year ago

A in my opinion slightly more elegant solutiin would be to use nodos.stream().map(NodoMoore.class::cast).collect(Collectors.groupingBy(NodoMoore::getSalida)), because it's better readable.

AFP_555 Over a year ago

Hello, I'm new to this "::" thing. Why is it more elegant?

andred Over a year ago

NodoMoore.class::cast is equal to p -> NodoMoore.class.cast(p). Using the ::-operator, you are implicitly either filling the parameters of the method or, if the method is non-static but supplied without an instance, giving the instance the method should run on. E.g. FooFactory::createFrom vs. Foo::toString.

andred Over a year ago

Ups, i got drifted slightly away from your question. It's more elegant, because it's more readable, even for non-Java geeks.

DaoWen · Accepted Answer · 2016-02-22 07:23:13Z

6

Rather than using a method handle (Nodo::getSalida), you can use a lambda and do the cast there instead:

Map<String, List<Nodo>> map = nodos.stream()
    .collect(Collectors.groupingBy(x->((NodoMoore)x).getSalida());

answered Feb 22, 2016 at 7:23

DaoWen

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3 Comments

AFP_555 Over a year ago

Amazing, I had no idea I could do this. Thanks!

DaoWen Over a year ago

Note that this answer and Eran's answer have different return types. You'd pick one approach or the other based on your desired return type.

AFP_555 Over a year ago

This was the desired output. I couldn't work with List<NodoMoore> because of the interface.

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