Java – C++ Namespaces, comparison to Java packages

cjavanamespacespackages

I've done a bunch of Java coding recently and have got used to very specific package naming systems, with deep nesting e.g. com.company.project.db. This works fine in Java, AS3/Flex and C#. I've seen the same paradigm applied in C++ too, but I've also heard that it's bad to view C++ namespaces as direct counterparts to Java packages.

Is that true, and why? How are namespaces/packages alike and different? What problems are likely to be seen if you do use deep nested namespaces?

Best Answer

In C++ namespaces are just about partitioning the available names. Java packages are about modules. The naming hierarchy is just one aspect of it.

There's nothing wrong, per-se, with deeply nested namespaces in C++, except that they're not normally necessary as there's no module system behind them, and the extra layers just add noise. It's usually sufficient to have one or two levels of namespace, with the odd extra level for internal details (often just called Details).

There are also extra rules to C++ namespaces that may catch you out if overused - such as argument-dependent-lookup, and the rules around resolving to parent levels. WRT the latter, take:

namespace a{ namespace b{ int x; } }
namespace b{ string x; }
namespace a
{
  b::x = 42;
}

Is this legal? Is it obvious what's happening? You need to know the precendence of the namespace resolution to answer those questions.

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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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