Java: accessing private constructor with type parameters

Question

This is a followup to this question about java private constructors.

Suppose I have the following class:

class Foo<T> {     private T arg;     private Foo(T t) {         // private!         this.arg = t;     }         @Override     public String toString() {         return "My argument is: " + arg;     }    } 

How would I construct a new Foo("hello") using reflection?

ANSWER

Based on jtahlborn's answer, the following works:

public class Example {     public static void main(final String[] args) throws Exception {         Constructor<Foo> constructor;         constructor = Foo.class.getDeclaredConstructor(Object.class);         constructor.setAccessible(true);         Foo<String> foo = constructor.newInstance("arg1");         System.out.println(foo);     }    } 

Answer 1

Make sure you use getDeclaredConstructors when getting the constructor and set its accessibility to true since its private.

Something like this should work.

Constructor<Foo> constructor= (Constructor<Foo>) Foo.class.getDeclaredConstructors()[0]; constructor.setAccessible(true);  Foo obj = constructor.newInstance("foo");  System.out.println(obj); 

Update

If you want to make use of getDeclaredConstructor, pass Object.class as an argument which translates to a generic T.

Class fooClazz = Class.forName("path.to.package.Foo"); Constructor<Foo> constructor = fooClazz.getDeclaredConstructor(Object.class); constructor.setAccessible(true);  Foo obj = constructor.newInstance("foo");  System.out.println(obj); 

Answer 2

you would need to get the class, find the constructor which takes a single argument with the lower bound of T (in this case Object), force the constructor to be accessible (using the setAccessible method), and finally invoke it with the desired argument.