Re: Java generics and type erasure
In article
<9d4c2b16-beb5-40b1-87a2-f03e971efeed@k17g2000vbn.googlegroups.com>,
Marcin Pietraszek <m.pietraszek@gmail.com> wrote:
Some time ago I've encountered stange behaviour while using generics,
small example is provided in gist:
https://gist.github.com/977599
Anybody could explain me why in second example (line commended with
"compilation failure") compilation fails?
You neglected to specify the type parameter for foo2, specified in the
declaration Foo<T>. Without the actual type, <Boolean>, the compiler
can only infer that get() returns Object, as would have been the case
prior to generics:
import java.util.*;
public class Foo<T> {
private Map<String, Integer> bar = new HashMap<String, Integer>();
public static void main(String... args) {
Foo<Boolean> foo1 = new Foo<Boolean>();
Integer x1 = foo1.bar.get("x"); // ok
Foo<Boolean> foo2 = new Foo<Boolean>();
Integer x2 = foo2.bar.get("x"); // compilation failure
}
Do you know any detailed description on how and when type erasure
works in java?
"All of these parameterized types share the same class at runtime."
<http://java.sun.com/docs/books/jls/third_edition/html/classes.html#8.1.2>
See also, Bloch, ch. 5:
<http://java.sun.com/docs/books/effective/>
--
John B. Matthews
trashgod at gmail dot com
<http://sites.google.com/site/drjohnbmatthews>