I have a variable that must meet two conditions, and I want to set them in the definition
I know that I can define either condition with an individual variable, like in any of these examples
private Class<? extends A> variable; //or
private A variable; //or
private Class<? extends B> variable; //or
private B variable;
But is there a way to have the variable meet both conditions?
I was hoping for something like this
private Class<? extends A implements B> variable;
But I can't find any way to do this without typecasting when I need to call it or storing multiple copies of it
You can declare type parameters that have multiple bounds, such as:
public static <T extends A & B> void test(Class<T> clazz)
But you cannot declare a variable that has multiple bounds:
private Class<? extends A & B> variable; // doesn't work
You can create an abstract
class C
that extends A
and implements B
, so that only one bound is required.
abstract class C extends A implements B {}
Then:
private Class<? extends C> variable;
While Java does not directly support intersection types like A&B
, such types do appear in type parameter bounds and capture conversions. We can express A&B
with a layer of abstraction.
public class ValueAB<T extends A&B>
{
public final T v;
// constructor ...
}
public class ClassAB<T extends A&B>
{
public final Class<T> clazz;
// constructor ...
}
Instead of A&B, Class<? extends A&B>
A&B, Class<? extends A&B>
, we use wrappers ValueAB, ClassAB
ClassAB<?> clazz = new ClassAB<>(Foo.class);
ValueAB<?> value = new ValueAB<>(clazz.c.newInstance());
value.v.methodOfA();
value.v.methodOfB();
This solution would require a wrapper for each combination of As and Bs.
Another solution is to use only A
as type parameter bound; B
will be supplied by wildcard bound. This is probably better if you need to express multiple A&B1, A&B2, ...
types at use site.
public class ValueA<T extends A>
{
public final T v;
...
}
public class ClassA<T extends A>
{
public final Class<T> c;
...
}
---
ClassA<? extends B> clazz = new ClassA<>(Foo.class);
ValueA<? extends B> value = new ValueA<>(clazz.c.newInstance());
If it's confusing how wildcard works in these cases, see my article on wildcard
A 3rd solution is free of A or B at declaration site of wrappers; the use site provides A and B.
public class Value<T extends S, S>
{
public final T v;
...
}
public class Clazz<T extends S, S>
{
public final Class<T> c;
...
}
---
Clazz<? extends A, B> clazz = new Clazz<>(Foo.class);
Value<? extends A, B> value = new Value<>(clazz.c.newInstance());
This is however probably too confusing.
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