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);