On Thu, Nov 6, 2008 at 1:49 PM, Jess Holle <[EMAIL PROTECTED]> wrote:
> Viktor Klang wrote: > > Actually guys, referential equality is kind of retarded, so the > > Map<String,Integer>.class == Map<Foo,Bar>.class > > example is bad. > > However, > > Map<String,Integer>.class.equals(Map<Foo,Bar>.class) SHOULD return false > > No, it MUST not -- else loads of things break. > If you want to make an omelet, you've got to break some eggs... > > > Further == and .equals MUST give the same results for Class objects -- else > loads of things break. > > The "ConcreteClass" noted below would not be a Class -- it would be a new > API. > Yeah, I realize that, that's why I'm moving away from Java. > > > -- > Jess Holle > > On Thu, Nov 6, 2008 at 12:38 AM, Jess Holle <[EMAIL PROTECTED]> wrote: > >> Peter Becker wrote: >> >> I think you probably want something like: >> >> Object -> ConcreteClass(String,Foo) -> GenericClass >> >> which adds the extra class instances, but they should be very small >> and not too many. >> >> Alternatively you could do: >> >> Object(String,Foo) -> GenericClass >> >> but then you'd have to store the type parameters on each object, which >> is probably much more expensive in total. The former approach seems to >> also match the type model better. >> >> >> Yes, I was thinking something along the lines of the former. >> >> On Thu, Nov 6, 2008 at 8:22 AM, Jess Holle <[EMAIL PROTECTED]> <[EMAIL >> PROTECTED]> wrote: >> >> >> Peter Becker wrote: >> >> Sorry: I missed that parameter in your method. >> >> Wouldn't adding this information lead to a potential explosion of >> Class instances at runtime similar to C++ templates? OTOH: there are >> only a limited number of type parameters you'd actually use in your >> code, so it is probably not too bad -- after all we wouldn't copy the >> whole code as C++ does, just get a construct refering to the generic >> class version and storing the type parameters. >> >> >> Avoding this explosion is a benefit of erasure as well. Dealing with C++ I >> quickly had dozens and dozens of copies of the same (sizable) object code >> all due to instantiation with different types -- even where the usage of the >> types in question (e.g. char*, void*, int*, Foo*, etc, in a vector<>) ended >> up being 100% equivalent from an object code perspective. I don't want to >> go near that sort of issue again. >> >> That said, I see no reason to have separate Class objects for >> Map<String,Foo> and Map<Bar,Baz>. This would lead to bloat and >> incompatibility. >> >> Rather one could have something like a "GenericTypesMap", ala: >> >> For class Map<K,V>, Map<String,Foo> would have a GenericTypesMap of >> {K->String,V->Foo} >> >> GenericTypesMap's could be shared across all instances which use the same >> instantation types and be weakly referenced by them or some such. >> >> I'm clearly just throwing together a strawman here, but the idea is to have >> a separate chunk of runtime data that spells out the generic types used by >> an instance without (1) breaking of existing Class contracts, explicit or >> implicit, (2) resulting in duplication of Class objects or other bloat, or >> (3) breaking interoperability between new and old code. >> >> -- >> Jess Holle >> >> On Thu, Nov 6, 2008 at 7:47 AM, Jess Holle <[EMAIL PROTECTED]> <[EMAIL >> PROTECTED]> wrote: >> >> >> getTypeParameters() will tell you that Map<K,V> is parameterized by K and V >> and if/how these are contrained by wildcards. >> >> It won't tell you that the Map passed to your method is a Map<String,Foo>, >> though. Map.class covers the generic notion of Map<K,V> -- it knows nothing >> about how a particular instance was parameterized and there's no such thing >> as a Map<String,Foo>.class in terms of this being any different than >> Map<K,V>. >> >> Peter Becker wrote: >> >> Like this: >> >> http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Class.html#getTypeParameters() >> >> <http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Class.html#getTypeParameters%28%29> >> >> ? >> >> Peter >> >> On Wed, Nov 5, 2008 at 2:35 PM, Jess Holle <[EMAIL PROTECTED]> <[EMAIL >> PROTECTED]> wrote: >> >> >> For the most part, Java 5 class files contain metadata indicating much of >> what the source file indicated as far as generics are concerned. This is >> certainly the case for field/method/class declarations. I'm not sure about >> local variable declarations, though. >> >> That said, once one has something like: >> >> void <T extends Foo> sort( List<T> list ) { ... } >> >> one can only determine that 'list' is parameterized by 'T', any >> extends/super constraints, etc. The body of sort() here has no other >> notions about T -- either in the class file or at runtime. That is >> erasure. List<A>.class == List<B>.class == List.class. This is necessary >> to keep the existing contracts and is a key benefit to erasure -- both in >> lack of class bloat and in preservation of existing contracts and >> compatibility. One could potentially have a special >> Class.getGenericTypeInfos(Object) utility that could seperately lookup this >> info, e.g. by having each object refer to both its class and its generic >> typing info -- rather than to just the class. When called by old, >> non-generic-savvy code the generic typing info would be null, of course. >> One could have the compiler do nifty bits with such a getGenericTypeInfos() >> utility so that one could do things like "new T[]" in sort -- throwing a >> runtime exception if the typing info is not present. This would be undoing >> erasure without blowing new/old code interoperability except where actually >> necessary. >> >> -- >> Jess Holle >> >> Christian Catchpole wrote: >> >> Here is my analysis of the situation. I could be wrong. But here >> goes.. >> >> When I got my copy of Java 5 my first question was, do generics really >> take the cast out of the equation? I disassembled the code to find >> the cast still exists. This implies that when you compile this.. >> >> HashMap<String,String> map = new HashMap<String,String>() >> String string = map.get(""); >> >> The generated code actually equates to this.. >> >> HashMap map = new HashMap() >> String string = (String)map.get(""); >> >> The class returned by map.getClass() does not know the map only >> contains Strings. It's actually the reference to the map which >> marshals the types. >> >> I did a quick test... >> >> HashMap<String,String> map1 = new HashMap<String,String>(); >> HashMap<Date,Date> map2 = new HashMap<Date,Date>(); >> >> System.out.println(map1.getClass() == map2.getClass()); >> >> true >> >> They use the same class and can't therefore hold the type information >> for both declarations. >> >> I can only assume this re-compiler the posse were talking about, scans >> the code for the actual cast / type check to determine the types. >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> > > > -- > Viktor Klang > Senior Systems Analyst > > > > > > > -- Viktor Klang Senior Systems Analyst --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "The Java Posse" group. 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