Java generics, get Class<T> of generic parameter

Question

I have an abstract class:

public abstract class RootProcessor<T> {     Class<T> clazz; } 

I need to fill ClassT clazz; with the children of RootProcessor - every child has its own T

I found only one solution, but it needs compiler argument -Xlint:unchecked

public RootProcessor(){     this.clazz = (Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass()).getActualTypeArguments()[0]; } 

Is this the best solution? Can we do the same without -Xlint:unchecked ?

Answer 1

The typesafe, but boilerplatey way to do this is to pass the Class<T> token "where the compiler can see it":

public abstract class RootProcessor<T> {     Class<T> clazz;      protected RootProcessor<T>(Class<T> clazz) {         this.clazz = clazz;     } }  public class FooProcessor extends RootProcessor<Foo> {     public FooProcessor() {         super(Foo.class);     } } 

If you're doing an unchecked cast but you "know what you're doing" and want the compiler to stop complaining, the correct approach would be localising the non-type-safe-but-you-know-they-work bits and using @SuppressWarnings:

public abstract class RootProcessor<T> {     Class<T> clazz;     { initClazz(); }      @SuppressWarnings("unchecked")     private void initClazz() {         // the usual verbiage you already have in your question         this.clazz = this.getClass().getGenericSuperclass().yadda().blah();     } } 

(I won't hold this against you :P)

Answer 2

There is a post of the same subject: Reflecting generics

And a class that implement it:TypeArgumentsUtils.java

An example is in the unit test.

So if you have this class:

public class BarProcessor extends RootProcessor<Bar> {     public BarProcessor() {     } } 

than you would get the first parameter with:

Class barClass = TypeArgumentsUtils.getFirstTypeArgument(         RootProcessor.class, BarProcessor.class);