Instantiating a generic class in Java [duplicate]

Answer 1

Here's a rather contrived way to do it without explicitly using an constructor argument. You need to extend a parameterized abstract class.

public class Test {   
    public static void main(String [] args) throws Exception {
        Generic g = new Generic();
        g.initParameter();
    }
}

import java.lang.reflect.ParameterizedType;
public abstract class GenericAbstract<T extends Foo> {
    protected T parameter;

    @SuppressWarnings("unchecked")
    void initParameter() throws Exception, ClassNotFoundException, 
        InstantiationException {
        // Get the class name of this instance's type.
        ParameterizedType pt
            = (ParameterizedType) getClass().getGenericSuperclass();
        // You may need this split or not, use logging to check
        String parameterClassName
            = pt.getActualTypeArguments()[0].toString().split("\\s")[1];
        // Instantiate the Parameter and initialize it.
        parameter = (T) Class.forName(parameterClassName).newInstance();
    }
}

public class Generic extends GenericAbstract<Foo> {
}

public class Foo {
    public Foo() {
        System.out.println("Foo constructor...");
    }
}

Answer 2

From https://stackoverflow.com/a/2434094/848072. You need a default constructor for T class.

import java.lang.reflect.ParameterizedType;

class Foo {

  public bar() {
    ParameterizedType superClass = (ParameterizedType) getClass().getGenericSuperclass();
    Class type = (Class) superClass.getActualTypeArguments()[0];
    try {
      T t = type.newInstance();
      //Do whatever with t
    } catch (Exception e) {
      // Oops, no default constructor
      throw new RuntimeException(e);
    } 
  }
}

Answer 3

Use The Constructor.newInstance method. The Class.newInstance method has been deprecated since Java 9 to enhance compiler recognition of instantiation exceptions.

public class Foo<T> {   
    public Foo()
    {
        Class<T> newT = null; 
        instantiateNew(newT);
    }

    T instantiateNew(Class<?> clsT)
    {
        T newT;
        try {
            newT = (T) clsT.getDeclaredConstructor().newInstance();
        } catch (InstantiationException | IllegalAccessException | IllegalArgumentException
            | InvocationTargetException | NoSuchMethodException | SecurityException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
            return null;
        }
        return newT;
    }
}

Answer 4

One option is to pass in Bar.class (or whatever type you're interested in - any way of specifying the appropriate Class<T> reference) and keep that value as a field:

public class Test {
    public static void main(String[] args) throws IllegalAccessException,
            InstantiationException {
        Generic<Bar> x = new Generic<>(Bar.class);
        Bar y = x.buildOne();
    }
}

public class Generic<T> {
    private Class<T> clazz;

    public Generic(Class<T> clazz) {
        this.clazz = clazz;
    }

    public T buildOne() throws InstantiationException, IllegalAccessException {
        return clazz.newInstance();
    }
}

public class Bar {
    public Bar() {
        System.out.println("Constructing");
    }
}

Another option is to have a "factory" interface, and you pass a factory to the constructor of the generic class. That's more flexible, and you don't need to worry about the reflection exceptions.