Why is this throwing null pointer exception?

Question

I have already asked one of the contradiction question here Why is this not throwing a NullPointerException?

But this is one of different type and behavior I want to know about, please look for my example below

package com;

public class test {

    public static void main(String[] args) {
        Abc abc = null;

        //Scenario 1
        System.out.println("ERROR HERE " + abc!=null?abc.getS1():""); //This is throwing null pointer exception 

        //Scenario 2
        String s1 = abc!=null?abc.getS1():"";
        System.out.println("This is fine " + s1);
    }
}

class Abc {
    String s1;

    public String getS1() {
        return s1;
    }

    public void setS1(String s1) {
        this.s1 = s1;
    }


}

So here Scenario 2 will work fine but why it is not working when I am trying it with other string concatenation in Scenario 1?

Due to order of evaluation of argument to `println()` in 1st case, `"ERROR HERE " + abc` is evaluated first and gives NPE. — Victor Sorokin, Dec 12 '13 at 16:00

score 5 · Answer 1 · answered Dec 12 '13 at 16:00

5

"ERROR HERE " + abc!=null?abc.getS1():""

is equivalent to

("ERROR HERE " + abc!=null)?abc.getS1():""

(which never evaluates to false and therefore you get NPE)

You meant:

"ERROR HERE " + (abc!=null?abc.getS1():"")

answered Dec 12 '13 at 16:00

Eyal Schneider

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3

Operator precedence rules are here: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html – James Cronen Dec 12 '13 at 16:02
1

The '+' operator is higher on the list than the ternary conditional operator '?:'. – James Cronen Dec 12 '13 at 16:03

Ross Drew · Answer 2 · 2013-12-12T16:09:12.820

2

You will need to separate it (with brackets) so that it's read as a whole separate statement

System.out.println("ERROR HERE " + (abc != null ? abc.getS1() : ""));

edited Dec 12 '13 at 16:09

answered Dec 12 '13 at 15:59

Ross Drew

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score 0 · Answer 3 · answered Dec 12 '13 at 16:19

+1 vote for Eyal and Ross.

After seeing both that answer and comment by Tenner, I have concluded that,

it is throwing nullPointerException because != operator priority is lower than +.

So compiler is calling toString method of Abc class and hence abc is null it is throwing nullPointerException.

Please correct me if I am wrong.

Why is this throwing null pointer exception?

3 Answers3