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From: Mike Schilling on 17 Mar 2010 15:07
Putting the following down on paper (OK, pixels) helped me to understand how
generics extend the Java type system. Since I haven't seen any texts that
look at things in quite this fashion, I thought I'd post it on the chance
that it would help others as well. Note that this only discusses object
types, not primitives, since it's describes the ability to refer to an
object of type A by a reference of type B, with B a supertype of A. It also
doesn't discuss conversions of the sort found in

float f = 2;

or

Integer i = 12;

In both cases, the entity on the right-hand side is converted to an entity
of a different type, quite different from what happens in

Object o = "hello";

in which the left hand-side is made to refer to precisely the entity that is
the right-hand side. OK, here we go:

The assignment statement in Java looks like

reference = expression;

This is legal whenever the type R of the reference is a supertype of the
type E of the expression, where "supertype" is defined so that, for all
types T, T is a supertype of itself. (As mentioned above, we're going to
ignore primitives and built-in conversion.) Before generics, the rules for
supertype were relatively simple:

1. T is a supertype of T
2. If T is a class other then Object, T's superclass is a supertype of T
3. Object is a supertype of all interfaces and array types
4. All interfaces that T implements or extends are supertypes of T
5. If T is a supertype of U:
T[] is a supertype of U[] (i.e. arrays are covariant)
6. If T is a supertype of U and U is a supertype of V, then T is a
supertype if V (i.e. "supertype" is transitive)

There rules are all pretty clear, even if you don't ordinarily think of them
that way.

Generics complicate this somewhat, by adding these extra rules.

If T is a supertype of U, and A is a supertype of B:
1. A<T> is a supertype of B<T>
2. A<T> is not a supertype of A<U> (i.e. generics are not covariant)
3. T is a supertype of <? extends T>
4. There is no type V for which <? extends T> is a supertype of V
5. <? super T> is a supertype of T
6. The only supertype of <? super T> is Object
7. A<? extends T> is a supertype of A<? extends U>
8. A<? super U> is a supertype of A<? super T>



From: Roedy Green on 17 Mar 2010 18:36
On Wed, 17 Mar 2010 12:07:59 -0700, "Mike Schilling"
<mscottschilling(a)hotmail.com> wrote, quoted or indirectly quoted
someone who said :

>t also
>doesn't discuss conversions of the sort found in
>
> float f = 2;
>
>or
>
> Integer i = 12;

For the first look under "widening", for the second look under
"autoboxing".
--
Roedy Green Canadian Mind Products
http://mindprod.com

Responsible Development is the style of development I aspire to now. It can be summarized by answering the question, �How would I develop if it were my money?� I�m amazed how many theoretical arguments evaporate when faced with this question.
~ Kent Beck (born: 1961 age: 49) , evangelist for extreme programming.
From: Mike Schilling on 17 Mar 2010 18:55
Roedy Green wrote:
> On Wed, 17 Mar 2010 12:07:59 -0700, "Mike Schilling"
> <mscottschilling(a)hotmail.com> wrote, quoted or indirectly quoted
> someone who said :
>
>> t also
>> doesn't discuss conversions of the sort found in
>>
>> float f = 2;
>>
>> or
>>
>> Integer i = 12;
>
> For the first look under "widening", for the second look under
> "autoboxing".



Yes, Roedy, I know what thty are, but they're not relevant to what I was
exploring.


From: Arne Vajhøj on 17 Mar 2010 20:09
On 17-03-2010 18:55, Mike Schilling wrote:
> Roedy Green wrote:
>> On Wed, 17 Mar 2010 12:07:59 -0700, "Mike Schilling"
>> <mscottschilling(a)hotmail.com> wrote, quoted or indirectly quoted
>> someone who said :
>>> t also
>>> doesn't discuss conversions of the sort found in
>>>
>>> float f = 2;
>>>
>>> or
>>>
>>> Integer i = 12;
>>
>> For the first look under "widening", for the second look under
>> "autoboxing".
>
> Yes, Roedy, I know what thty are, but they're not relevant to what I was
> exploring.

Most likely Roedy (as usual) did not read the entire message that he
was replying to.

Arne
From: Mike Schilling on 17 Mar 2010 21:44
Stefan Ram wrote:
> Arne Vajh�j <arne(a)vajhoej.dk> writes:
>> Most likely Roedy (as usual) did not read the entire message
>> that he was replying to.
>
> Mike posted some very helpful observations at the end of
> his post. But he started his post with distracting and
> boring preliminaries. Possibly Roady was not the only one
> who was not reading it to the end. Anyway, I thank Mike
> for sharing his insights!

You're welcome. (I think.)


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