Java Try Catch NullPointerException

Question

I was trying an example about try catch but it isn't working as I wanted. This is the code :

public static void main(String[] args)
{
    //This array's size is 5 and it has 4 Strings in it. 
    String[] array = new String[5];
    array[0] = "Hello";
    array[1] = "World";
    array[2] = "try catch";
    array[4] = "error";

    try
    {
        for(int i = 0 ; i<array.length ; i++)
            System.out.println("This array has: "+array[i]);

    } catch (NullPointerException e)
    {
        System.out.println("Null!!!");
    }
}

output is like that :

This array has: Hello

This array has: World

This array has: try catch

This array has: null

This array has: error

It supposed to enter catch block but it didn't. Any ideas?

What operation did you expect to throw the exception, and why? (Also, it would be appreciated if you could format your code carefully before posting - the indentation is all over the place here.) — Jon Skeet
– Jon Skeet, Commented Dec 15, 2014 at 18:40
array[3] is null and hence it prints that. Why should it throw an NPE? — TJ-
– TJ-, Commented Dec 15, 2014 at 18:40

rgettman · Accepted Answer · 2014-12-15 18:41:01Z

7

A NullPointerException occurs when a method is called on an object reference that is null or you attempt to access a null array. But that doesn't occur here. The array is initialized properly. Even though you didn't initialize array[3], the NPE doesn't occur here.

What does occur here is String conversion (JLS, Section 5.1.11).

If the reference is null, it is converted to the string "null" (four ASCII characters n, u, l, l).

A null String is converted to the String "null" when concatenated, so no NPE occurs.

answered Dec 15, 2014 at 18:41

rgettman

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M A · Accepted Answer · 2014-12-15 18:41:33Z

2

The elements of object arrays (all types except primitives) are initialized to null. When a null element is printed, the string literal "null" is displayed as shown in the below snippet of PrintStream#print used by System.out.println:

public void print(String s) {
    if (s == null) {
        s = "null";
    }
    write(s);
}

Note that it's a bad practice to catch NullPointerExceptions. If it were a checked exception, the compiler would have probably flagged that the exception can never occur.

answered Dec 15, 2014 at 18:41

M A

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Pelit Mamani · Accepted Answer · 2014-12-15 18:41:26Z

0

It's not supposed to get into the 'catch' block. there is no NullPointerException for REQUESTING array[i] even if it happens to hold null. You will get an exception if you try to do anything with this null, e.g. array[i].toLowerCase()

answered Dec 15, 2014 at 18:41

Pelit Mamani

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Semih Eker · Accepted Answer · 2014-12-15 18:43:31Z

0

The elements of array are initialized as null. If you want to get NullPointerException, try to use String methods like;

array[i].toLowerCase()

answered Dec 15, 2014 at 18:43

Semih Eker

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JEuvin · Accepted Answer · 2014-12-15 18:46:29Z

0

Try to do something with the string to create an error.

Do something like 'split(String regex)' this will cause the script to blow up cause you trying to split a null value. It will hit the catch portion of your script.

answered Dec 15, 2014 at 18:46

JEuvin

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