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From: Susan Calvin <s.calvin@usr.invalid>
Newsgroups: comp.lang.java.programmer
Subject: Re: Java generics and type erasure
Date: Wed, 25 May 2011 00:13:08 +0000 (UTC)
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On Tue, 24 May 2011 23:05:25 +0000, Ian Shef wrote:

> "The type of a constructor (§8.8), instance method (§8.8, §9.4), or non-
> static field (§8.3) M of a raw type C that is not inherited from its
> superclasses or superinterfaces is the erasure of its type in the
> generic declaration corresponding to C."

This seems silly to me. I mean, if you have a class that starts:

public class Foo <T> {
    public Map<String,Integer> m1;
    public Map<String,T> m2;

then the logical way to deal with a raw Foo is to treat it as if T were ? 
(or ? extends whatever its lower type bound was, or etc.), so in

Foo aFoo = new Foo();

aFoo.m1 would have compile-time type Map<String,Integer> (since no matter 
what T was, it would be a Map<String,Integer>) and aFoo.m2 would have 
compile-time type Map<String,?> -- disallowing puts and having gets have 
type ?, assignable only to Object, but still recognizing the key type as 
String.

The only reason I can think of for not doing this (the logic seems simple 
enough to implement) is that it turned out doing so would break legacy 
code that used util collections' raw types. Was that why? Perhaps there 
should be a compile flag that turns on the legacy-compatible behavior for 
use when compiling 1.4 and older sources, but which is off by default? 
Though I guess it's too late for that now. :P