Printing out strings in an array while not printing the null, Java

Question

When I run my method to print out the array, it will give me a NullPointerException and I have no idea how to get it to just print what I want. Let's say I want Array[0] to be "Sally" and Array[3] to be "Jeff", I want it to print out 0, Sally 3, Jeff. While leaving all the null slots of the array alone.

Here's my code:

public void printAll()
{       
    for(int i = 0; i <= 10; i++)
    {
        if(seats[i].equals(null))
        {
            System.out.print(i + " , " + seats[i] + "\n");              
        }

    }
}

Any help would be greatly appreciated, if I'm being too vague I can reply with more details.

null has no equals, use ==

user180100
– user180100

2014-02-19 06:34:50 +00:00
Commented Feb 19, 2014 at 6:34 — user180100
– user180100, Commented Feb 19, 2014 at 6:34

paxdiablo · Accepted Answer · 2014-02-19 06:34:49Z

1

You need:

if (seats[i] != null)

Using equals will cause a de-reference, which is exactly what you're trying to avoid.

answered Feb 19, 2014 at 6:34

paxdiablo

888k243 gold badges1.6k silver badges2k bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Scary Wombat · Accepted Answer · 2014-02-19 06:36:31Z

1

You should not use:

if(seats[i].equals(null))

That condition will itself throw a NPE. You should use:

if(seats[i] != null)

edited Feb 19, 2014 at 6:36

Scary Wombat

44.9k6 gold badges38 silver badges67 bronze badges

answered Feb 19, 2014 at 6:35

Rohit Jain

214k45 gold badges419 silver badges534 bronze badges

Comments

Ankur Shanbhag · Accepted Answer · 2014-02-19 06:35:27Z

0

Change if(seats[i].equals(null)) to if(seats[i] != null) or if(seats[i] == null) to avoid NPE.

Because, if seats[i] is null you cannot call methods on null references.

answered Feb 19, 2014 at 6:35

Ankur Shanbhag

7,8042 gold badges31 silver badges38 bronze badges

Comments

ErstwhileIII · Accepted Answer · 2014-02-19 06:37:48Z

0

You may want to use the following: (you were checking equality of objects, not whether the array item was null.

public void printAll()
{   
...
    if (seats != null) {
        for (int i=0; i < seats.size(); i++) {
            if (seats[i] != null) {
                System.out.println(i + ", " + seats[i]);  
            }
        }
    }  

}

answered Feb 19, 2014 at 6:37

ErstwhileIII

4,8722 gold badges25 silver badges37 bronze badges

Comments

TheBrenny · Accepted Answer · 2014-02-19 06:37:50Z

0

if(seats[i].equals(null)) {
  System.out.print(i + " , " + seats[i] + "\n");              
}

This is causing your null pointer exception because seats[i] is equal to null. When you have a null object, you can't reference it. At all. This is logical because if you have nothing, then you can't ask what it has.

To fix this, the best way to check if something is equal to null is by going if(Object == null) and if it ISN'T null then, if(Object != null). These are very common through any code you may have.

Jarod.

answered Feb 19, 2014 at 6:37

TheBrenny

5373 silver badges19 bronze badges

Comments

pretty_monk · Accepted Answer · 2014-02-19 07:00:08Z

0

public static void printAll(){

  String[] seats={"Jerry","chen","Jack"};
  for(int i = 0; i <seats.length; i++)
  {

if(!seats[i].equals(null))
    {
        System.out.print(i + " , " + seats[i] + "\n");              
    }

}

answered Feb 19, 2014 at 7:00

pretty_monk

212 bronze badges

Collectives™ on Stack Overflow

Printing out strings in an array while not printing the null, Java

6 Answers 6

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related