Eliminate type parameter of java generics

Question

The code:

interface Property<T>
{
    T get();
}

class BoolProperty implements Property<Boolean>
{
    @Override
    public Boolean get()
    {
        return false;
    }
}
class StringProperty implements Property<String>
{
    @Override
    public String get()
    {
        return "hello";
    }
}
class OtherStringProperty implements Property<String>
{
    @Override
    public String get()
    {
        return "bye";
    }
    public String getSpecialValue()
    {
        return "you are special";
    }
}

is used by my class:

class Result<P extends Property<X>, X>
{
    P p;
    List<X> list;
}

As you see it has two type parameters P and X. Despite of that the X can always be deduced from P but the language requires me to supply both:

Result<BooleanProperty, Boolean> res = new Result<BooleanProperty, Boolean>();

Is there any trick to get rid of the X type parameter? I want just use

Result<BooleanProperty> res = new Result<BooleanProperty>();

Also, I don't want lose type information and use it as:

Result<OtherStringProperty> res = new Result<OtherStringProperty>();
String spec = res.p.getSpecialValue();
String prop = res.list.get(0);

Answer 1

I would change Result class to be something like

class Result<X> {
    Property<X> property;
    List<X> list;
}

I don't think the compiler can infer X from Property, as your Result class is waiting two definitions for the two generics.