CODEWITHSUNDEEP
javaarraysgenerics
Mohamed
Mohamed

Reputation: 193

Java Generics Creating Array from Class

I have a hierarchy where Square, Triangle and Circle all extend from Shape. I have a working method:

public void someMethod() {
   File file = new File("File_with_squares");
   ThirdPartyClass foo = new ThirdPartyClass();
   Square[] squares = foo.someMajicMethod(Square[].class,file);
   for (Square square: squares) 
      square.draw();

}

Now I want to make this method generic so that it can accept any shape. I want to be able to call it someMethod(Triangle.class,new File("File_with_triangles") or someMethod(Circle.class, new File("File_with_circles"). I am trying like this:

public void someMethod(Class<? extends Shape> type, File shapeFile) {
   ThirdPartyClass foo = new ThirdPartyClass();
   #### What goes here??? ####
   for (Shape shape: shapes)
       shape.draw();
}

What should be there at #### What goes here??? #### ???

Upvotes: 6

Views: 7645

Answers (3)

avh4
avh4

Reputation: 2655

Assuming ThirdPartClass.someMajicMethod has a signature something like this:

public <T> T someMajicMethod(Class<T> class1, File file);

Then you should be able to do something like this:

public void someMethod(Class<? extends Shape> type, File shapeFile) {
    ThirdPartyClass foo = new ThirdPartyClass();

    @SuppressWarnings("unchecked")
    Class<? extends Shape[]> arrayType = 
        (Class<? extends Shape[]>) Array.newInstance(type, 0).getClass();
    assert Shape[].class.isAssignableFrom(arrayType);

    Shape[] shapes = foo.someMajicMethod(arrayType, shapeFile);

    for (Shape shape: shapes)
        shape.draw();
}

So if you call someMethod(Triangle.class, file), then arrayType will be Triangle[].class in the call to someMajicMethod.

Though you may find it simpler to have someMethod take the array type as a parameter instead of the element type so you can avoid that step.

Upvotes: 6

Jesse Webb
Jesse Webb

Reputation: 45243

Shape[] shapes = foo.someMajicMethod(type, file);

If foo is a third-party class, I assume you don't control the API of it. I assumed it has the appropriate method signature to handle the line I've written but there is no way for me to be certain without more information about that class.

If this doesn't work, what is the problem?

Upvotes: 3

Omnaest
Omnaest

Reputation: 3096

Perhaps Array.newInstance(..) is of interest for you

Upvotes: 6

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