NullPointerException or not

[Janet Smith]

Hi,

I'm currently working my way through the tests on the CD that comes with the OCA/OCP Java SE7 Programmer I & II Study guide. I got a question that contained the following code and was asked to predict the outcome:

import java.util.*;
class Sorter{
	public static void main(String[] args){
		ArrayList<String> sa = new ArrayList<String>();
		sa.add("a");
		sa.add("b");
		sa.add("c");
		String sb[] = new String[3]; 
		for(int x = 1; x < sa.size(); x++)
			sb[x] = sa.get(x);
		for(String s: sb)
			System.out.print(s + " ");

	}
}

I got this question wrong because I thought I would get a NullPointerException as the first for loop populates elements 1 and 2 of the array, but the second for loop prints all elements, including the one at position 0, which the code never populated.

The correct answer is that the output is "null b c"

I looked through the book and chapter 5 says "If you actually try to use that null reference by, say applying the dot operator to invoke a method on it, you'll get the infamous NullPointerException"

So I can do a System.out.print on sb[0] which will print "null", but if I tried to call sb[0].toLowerCase() I would get a NullPointerException?

I was wondering if someone could explain a little about this? I had it in my head that any reference to sb[0] would throw a NullPointerException so I'm interested to know what will and what won't throw it.

[Henry Wong]

Janet Smith wrote:
I was wondering if someone could explain a little about this? I had it in my head that any reference to sb[0] would throw a NullPointerException so I'm interested to know what will and what won't throw it.

If you dereference a null reference, meaning use it to access an instance variable or call an instance method, you will get a NullPointerException exception. It is perfectly fine to pass a null reference around. It is also fine to use a null reference to access static variable or static methods. In the case of the println() example, you simply passed the reference to the method (which is fine) -- and luckily, that method has code that checks to see if the reference is null, and if so, replaces it will a string that is set to "null".

Henry

[Winston Gutkowski]

Henry Wong wrote:and luckily, that method has code that checks to see if the reference is null, and if so, replaces it will a string that is set to "null".

@Janet: Specifically, println() uses String.valueOf(Object) to determine what string to print out, and if you look at the docs for that method, you'll read precisely the behaviour you're seeing.

HIH

Winston

[Janet Smith]

Thanks both of you for those explanations.

[Roel De Nijs]

Winston Gutkowski wrote:@Janet: Specifically, println() uses String.valueOf(Object) to determine what string to print out, and if you look at the docs for that method, you'll read precisely the behaviour you're seeing.

I have to nitpick a little bit about that statement This would have been 100% accurate if the line of code at line12 would have been

System.out.print(s);

In this case the overloaded version of the print() method taking an Object parameter would be executed. And the implementation of this method uses indeed String.valueOf(Object) to create a string representation of the Object parameter. And that method will (as Henry mentioned) check for null; it returns "null" in that case, object.toString() otherwise (that illustrates why the null check is so important). So if you would use the statement System.out.print(s); (on line12) in the above code snippet, the output will be "nullbc".

But the above code snippet uses a different statement

System.out.print(s + " ");

In this case concatenating s with " " happens first and then the result (a String object) will be passed to the overloaded version of the print() method taking an Object parameter. So now a new question arises: why does the concatenation s + " " not throw a NullPointerException? Luckily the answer is very simple (and already discussed). The same method String.valueOf(Object) is used here as well. So s + " " is executed as String.valueOf(s) + " ". And that's why no NPE will be thrown. So if you would use the statement System.out.print(s); (on line12) in the above code snippet, the output will be "null b c ".

And here is a final note. You know two different ways to concatenate strings: either use the concatenation operator + or use the concat(String) method of the String class. Although both ways will concatenate, there are a few important differences:

if a is null, then a.concat(b) throws a NullPointerException but a + b will not; so if you would use the statement System.out.print(s.concat(" ")); (on line12) in the above code snippet, a NullPointerException will be thrown

if b is null, then b.concat(a) throws a NullPointerException but a + b will not; so if you would use the statement System.out.print(" ".concat(s)); (on line12) in the above code snippet, a NullPointerException will be thrown; using statement System.out.print(" " + s); (on line12) in the above code snippet will result in the output " null b c"

the concat() method only accepts String values while with the + operator you can append all other types (primitive data types and reference types) as well; using statement System.out.print(s.concat(1)); (on line12) in the above code snippet results in a compiler error; if you would use the statement System.out.print(s + 1); (on line12) in the above code snippet, the output will be "null1b1c1"

Hope it helps!
Kind regards,
Roel

[Winston Gutkowski]

Roel De Nijs wrote:So now a new question arises: why does the concatenation s + " " not throw a NullPointerException?...

Good point, and nice explanation. Have a cow.

Winston