Java: Preferences API vs. Apache Commons Configuration

apache-commonsconfigurationjava

I need to allow the user to store/load an arbitrary number of lists of objects (assume they are Serializable). Conceptually I want a data model like

class FooBean { /* bean stuff here */ }

class FooList {
    final private Set<FooBean> items = new HashSet<FooBean>();

    public boolean add(FooBean item) { return items.add(item); }
    public boolean remove(FooBean item) { return items.remove(item); }
    public Collection<FooBean> getItems() { 
       return Collections.unmodifiableSet(items); 
    }
}

class FooStore {
    public FooStore() {
       /* something... uses Preferences or Commons Configuration */ 
    }
    public FooList load(String key) {
       /* something... retrieves a FooList associated with the key */ 
    }
    public void store(String key, FooList items) { 
       /* something... saves a FooList under the given key */
    }
}

Should I use the Preferences API or Commons Config? What's the advantages of each?

Best Answer

Well, commons-configuration is, like many of apache project, an abstraction layer allowing one to seemlessly use preferences, an ldap store, properties files, and so on. So, your question could be rewritten as is : Will you need to change the format you use to store your preferences ? If no, java preferences are the way to go. Elsewhere, consider the portability of commons configuration.

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Java – What are the differences between a HashMap and a Hashtable in Java

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 Java “pass-by-reference” or “pass-by-value”

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

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