|
Lim,I've
IIRC you are fairly new to Java. I am by no means a Java expert, but
taught myself and been there so I thought I would give you a few tips.ArrayList<String>();.
Firstly, always try to use the most generic type possible. So, your
ArrayList should be a List, e.g. List<String> rs = new
with that
With a number as small as 20 there is no real need to construct it
size.but
If it were me I would probably also not return an array of strings,
return a collection of type string. If you are worried about it beimmutable
look at Collections.unmodifiableList().need
Now, all that being said if you still need to return an array all you
to do is rs.toArray(new String[0]). This has to do with generics. TheThe
default List.toArray() method will only return an array of objects.
List.toArray(T[]) is used to return the generic type. Without lookingat the
code I couldn't tell you why it's designed like that, but Joshua Blochhad a
lot to do with the Collections library, so I trust there may be a good[Ljava.lang.Object;
reason.
At any rate I hope this helps. Let me know if you want any clearer
clarification.
--
James R. Perkins
http://twitter.com/the_jamezp
On Tue, Sep 15, 2009 at 06:49, hockchai Lim
<lim.hock-chai@xxxxxxxxxxxxxxx>wrote:
I'm getting a cast error "java.lang.ClassCastException:
resolveincompatible with [Ljava.lang.String;" and not sure why and how to
InputStreamReader(process.getInputStream()));it. Help. . . ..
ArrayList<String> rs = new ArrayList<String>( 20 );
String str;
bin= new BufferedReader(new
error:
while ((str = bin.readLine()) != null) {
rs.add(str);
}
return (String[]) rs.toArray(); // =======> Getting the casting
(JAVA400-L)
}
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