There are several differences between HashMap
and Hashtable
in Java:
Hashtable
is synchronized, whereas HashMap
is not. This makes HashMap
better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable
does not allow null
keys or values. HashMap
allows one null
key and any number of null
values.
One of HashMap's subclasses is LinkedHashMap
, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap
for a LinkedHashMap
. This wouldn't be as easy if you were using Hashtable
.
Since synchronization is not an issue for you, I'd recommend HashMap
. If synchronization becomes an issue, you may also look at ConcurrentHashMap
.
Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In the example above aDog.getName()
will still return "Max"
. The value aDog
within main
is not changed in the function foo
with the Dog
"Fifi"
as the object reference is passed by value. If it were passed by reference, then the aDog.getName()
in main
would return "Fifi"
after the call to foo
.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In the above example, Fifi
is the dog's name after call to foo(aDog)
because the object's name was set inside of foo(...)
. Any operations that foo
performs on d
are such that, for all practical purposes, they are performed on aDog
, but it is not possible to change the value of the variable aDog
itself.
For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
Best Answer
Java SE = Standard Edition. This is the core Java programming platform. It contains all of the libraries and APIs that any Java programmer should learn (java.lang, java.io, java.math, java.net, java.util, etc...).
Java EE = Enterprise Edition. From Wikipedia:
In other words, if your application demands a very large scale, distributed system, then you should consider using Java EE. Built on top of Java SE, it provides libraries for database access (JDBC, JPA), remote method invocation (RMI), messaging (JMS), web services, XML processing, and defines standard APIs for Enterprise JavaBeans, servlets, portlets, Java Server Pages, etc...
Java ME = Micro Edition. This is the platform for developing applications for mobile devices and embedded systems such as set-top boxes. Java ME provides a subset of the functionality of Java SE, but also introduces libraries specific to mobile devices. Because Java ME is based on an earlier version of Java SE, some of the new language features introduced in Java 1.5 (e.g. generics) are not available.
If you are new to Java, definitely start with Java SE.