Martin Wolf - All Comments

re: Advanced Java 5 Generics

Sham — Wed, 10 Mar 2010 14:16:35 GMT

That was a great piece of demonstration

re: I want coroutines!

Daniel Sperry — Wed, 17 Feb 2010 15:33:58 GMT

Here an approach to coroutines in java http://jyield.googlecode.com/

Check the examples, there is one that is same from the msdn c# yield reference but written in java.

re: I want coroutines!

Marcin Rzeźnicki — Mon, 05 Oct 2009 16:28:23 GMT

If you are interested in java coroutines you might find my project useful: code.google.com/.../coroutines

Recent Links Tagged With "coroutines" - JabberTags

Recent Links Tagged With "coroutines" - JabberTags — Fri, 30 Jan 2009 11:54:33 GMT

PingBack from http://www.jabbertags.com/popular/coroutines

re: Christmas Algorithms

Raimondb — Sun, 03 Dec 2006 23:51:25 GMT

Doesn't the shuffle part already fill the fairness requirement? So that the sequence parts does not really matter anymore in terms of hanving random combinations?

blog.leenarts.net » Gift drawing code

blog.leenarts.net » Gift drawing code — Thu, 30 Nov 2006 22:14:43 GMT

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re: Advanced Java 5 Generics

Anonymous — Tue, 26 Apr 2005 18:39:00 GMT

Oops, should have replaced the Collection interface with the List to be able to use the get() method, but I hope you get the point. :)

re: Advanced Java 5 Generics

Anonymous — Tue, 26 Apr 2005 18:36:00 GMT

Very interesting article. The compiler can (and will) make some assumptions about wildcard types though. Perhaps a useful addition to the example:

Collection<? extends Animal> dogs = new ArrayList<Dog>();
dogs.add(new Snake("Hissy")); // This won't compile..
dogs.add(null); // This will work, though.
Animal animal = dogs.get(0); // ** This will work as well!

"? extends Animal" means "it can be any type, as long as it will fit into an Animal", so assigning the wildcard type to an Animal is safe, and the compiler will allow this. If you write this on the other hand:

Collection<? super Dog> dogs = new ArrayList<Animal>();
dogs.add(new Snake("Hissy")); // This won't compile again..
dogs.add(new Dog("Roger")); // ** This will work though!
Animal animal = dogs.get(0); // ** won't work without a cast!

"? super Dog" means "it can be any type, as long as I can fit a Dog in it", so assigning a Dog to the wildcard type is safe. On the other hand, this wildcard type has no meaning when assigning from it, because it could be any supertype of Dog, including Object.

An article about Java 5 Generics

TrackBack — Thu, 14 Apr 2005 17:18:00 GMT