Vector versus Hashtable and Related Questions

From:
Russ <pyth7@verizon.net>
Newsgroups:
comp.lang.java.help
Date:
Fri, 26 Nov 2010 22:38:36 -0800 (PST)
Message-ID:
<8a57b371-a7df-4fee-9e50-59c0466b3366@c39g2000yqi.googlegroups.com>
Hello All,
This question has been probably "hashed" over many times and googling
the question did answer the question on one hand but also raised
others.
The initial question is: Is a Vector or Hashtable efficient for
searching for clients in say a server to deliver a message? Looking at
this link:
http://www.devx.com/tips/Tip/25941

It says: "Searching through a Vector is O(n), which means that given n
elements, you may have to look at every element before finding what
you are looking for. Unless you use a value as its own key, a
Hashtable will be equally poor."

However that "Unless" creates questions, because looking at some old-
school code (Java 1) from an open-source server where the coder (who
has far more skill than myself) wrote the following method where
"users" is a Vector:

 synchronized VUser findUser(String name) {
    for (int i = 0; i < users.size(); i++) {
        VUser user = (VUser) users.elementAt(i);
        if (user.name.equals(name)) {
        return user;
        }
    }
    return null;
    }

Why did the coder do it that way? Isn't "users" as a Hashtable more
efficient?

synchronized VUser findUser(String name) {
        VUser user = (VUser) users.get( name );
        return user;
         }

It seems obvious that if "users" is a Hastable and once the instance
of VUser is created to simply map the Value to its own key of a
protected String name within the Value:

VUser VU = new vUser( String name );
users.put( VU.name, VU );

Also, why use "synchronized" on the findUser method when both Vectors
and Hashtables are already synchronized?

Finally, suppose I create an instance of a class and put that instance
in two different Hashtables
or Vectors. If I change a variable in that instance in one Hashtable
won't the same instance in the other Hashtable be changed or updated
as well? Common sense says it is the same allotment of memory so yes,
but I am just trying to be certain.

Thanks!
Russ Kinter

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