Java generic definition in a static method

Question

I have the following problem when using java generic. Basically I need to do a conversion from an list of subtype to a list of its supertype, and I defines a method like this:

public static <T> List<T> getList(List<? extends T> input) {
    ArrayList<T> list = new ArrayList<T>();
    list.addAll(input);
    return list;
}

However I'm not sure how I can invoke this method? Given type A, B extends A:

List<A> alist = getList(List<B> blist);

has a mismatch type, because in this case I only supply B's type information to the method invocation.

How can I invoke the method with the type information A supplied?

Thanks.

ColinD · Accepted Answer · 2011-07-27 02:48:07Z

5

List<B> blist = ...
List<A> alist = ClassName.<A>getList(blist);

Note that you could also just use ArrayList's constructor here:

List<A> alist = new ArrayList<A>(blist);

answered Jul 27, 2011 at 2:48

ColinD

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5 Comments

Eng.Fouad Over a year ago

I didn't know that you can do ClassName.<A>getList(blist) in java

Greg Mattes Over a year ago

@ColinD Shouldn't it be ClassName.<A, B>getList(blist)? Specifying both generic types? I seem to need that on my Java 6 compiler at least...

Greg Mattes Over a year ago

I think that the "just call the constructor" route is the way to go.

ColinD Over a year ago

@Greg: No, the method only has one generic type parameter, T, and as such can only accept one type argument, A in this case. A matters because it's the return type we want. B doesn't matter so much other than that it must be a subtype of A.

Greg Mattes Over a year ago

@ColinD: ah, yes, you're correct. I was playing around with some code for this question and at one point I had a method that had two type parameters, but both don't need to be named, one can be a wildcard, my bad, thanks (I still think that just calling the constructor is likely the best approach ;)

Jim Garrison · Accepted Answer · 2011-07-27 02:57:44Z

-1

How about

public static void <T, S extends T> List<T> getList(List<S> input)

answered Jul 27, 2011 at 2:57

Jim Garrison

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2 Comments

ColinD Over a year ago

That doesn't add anything over the List<? extends T> version unfortunately. Calls still fail to compile without specifying type arguments.

newacct Over a year ago

that's right: there's no point to add S if it's only used once

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