Java – the use of try-catch in setting a date value in java

date formatjavatry-catch

I'm a java newbie going through some training material, and this is one of the questions.

Given a valid DateFormat object df, and

Date d = new Date(0L);
String ds = "December 12, 2012";
//Insert correct code here

What updates d's value with date represented by ds?

A. d = df.parse(ds);
B. d = df.getDate(ds);
C. try{
        d = df.parse(ds);
       }
   catch(ParseException e){ };

The correct answer is C. Why is that so? What is the difference between A and C?

Best Answer

Because parse() can throw a ParseException and it is a checked Exception. Checked Exceptions must be handled by the calling code using a try-catch block or your code must declare that it can throw the Exception , by using a throws clause.

Checked exceptions are exceptions that the designers of Java feel that your programs absolutely must provide for, one way or another. Whenever you code a statement that could throw a checked exception, your program must do one of two things:

Catch the exception by placing the statement within a try statement that has a catch block for the exception.
Specify a throws clause on the method that contains the statement to indicate that your method doesn’t want to handle the exception, so it’s passing the exception up the line.

A better code would have been :

try{
    d = df.parse(ds);
   }
catch(ParseException e){   
   e.printStackTrace();
   // log the exception
   throw new RuntimeException(e);
}

Read this for more on Checked Exceptions.

Related Solutions

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

Related Solutions

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

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

Related Topic