| Is one class a superclass of another? [message #200996] |
Thu, 14 April 2005 20:42  |
Eclipse User |
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I'm trying to write a method that receives a Class object and later tries
to create an object of that class. For sanity check this object must be
derived from another class. Something like that:
class B extends A {};
void method(Class bClass)
{
// sanity check goes here
}
void anotherMethod()
{
method(B.class);
}
With all possible warnings on in 3.1M6 with JDK 5.0 it compiles fine. But
when I try to implement sanity check to see if bClass is actually a
subclass of A, I get all kinds of warnings.
If I use 'if (!bClass.isAssignableFrom(A.class))' I get warning 'Type
safety: The method isAssignableType(Class) belongs to raw type Class.
References to generic type Class<T> should be parametrized.'
If I parametrize method to look like 'void method(Class<A> bClass)', this
warning disappears, but I get an error when I try to invoke method() in
anotherMethod: 'The method method(Class<A>) is not applicable for the
arguments (Class<B>)'.
If instead of isAssignableForm() I use newInstance() and isInstance(), on
newInstance() line it gives me the same warning. Simply put, if I use any
Class method without parametrizing the bClass type, I get a warning. If I
parametrize it, I get an error.
Is there a right solution available or should I just go with the lesser of
two evils and have a warning there?
Thanks
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| Re: Is one class a superclass of another? [message #203167 is a reply to message #201012] |
Wed, 11 May 2005 13:38  |
| Eclipse User |
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Because java.lang.Class is now generic, the raw form is best avoided. The
simplest rewrite is to make the sanity method look like:
void method(Class<?> bclass) { ... }
But you could even write this to provide "static" type checking:
void method2(Class<? extends A> bclass) { /* do nothing */ }
In your example this will work, because the type of B.class is now Class<B>.
This works well with class literals which can be strongly typed, but what
about classes that are dynamically loaded? For example: the type of
Class.forName("B") is Class<?>, so this invocation would fail at compile
time:
method2(Class.forName("B"))
Instead you could use the new "asSubclass" method:
method2(Class.forName("B").asSubclass(A.class));
This allows you to have stronger typing at compile time, then get rid of the
sanity method, and replace all references to the raw Class with Class<?
extends A>. This has the added advantage of not having to cast the result
Class.newInstance(). For example:
Class<? extends A> clazz = Class.forName("B").asSubclass(A.class);
A newobj = clazz.newInstance();
"Genady" <eclipse@genady.org> wrote in message
news:d3nkur$ot9$1@news.eclipse.org...
>I think that "isAssignableFrom() is the right way to check subclassing". I
>used it and it works.
> If you use generics you can of course enforce subclassing in a more clean
> way.
>
> Genady Beryozkin
> http://www.genady.net/
>
>
>
> Aare Tali wrote:
>
>>
>>> If I use 'if (!bClass.isAssignableFrom(A.class))' I get warning 'Type
>>> safety: The method isAssignableType(Class) belongs to raw type Class.
>>> References to generic type Class<T> should be parametrized.'
>>
>>
>> Note to self: I had it backwards, 'if
>> (!A.class.isAssignableFrom(bClass))' didn't complain about anything.
>> Whether it works, I don't know yet.
>>
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