Null Pointer Exception in String Comparison

Question

while(true)
        {
            String input = "";
            try {
                input = in.readLine();

            } catch (IOException e1) {
                // TODO Auto-generated catch block
                System.out.println(e1 + "Exception occured when reading user input");
            }
            // Sleep
            try
            {
                Thread.sleep(USER_THROTTLE);
            }
            catch(Exception e)
            {
                System.out.println(toString()+" has input interrupted.");
            }

            if(input .equals("w")){action_event(input);}
            if(input .equals("a")){action_event(input);}
            if(input .equals("s")){action_event(input);}
            if(input .equals("d")){action_event(input);}

            if(input .equals("eat")){action_event(input);}
            if(input .equals("drink")){action_event(input);}
            if(input .equals("place")){action_event(input);}
            if(input .equals("swim")){action_event(input);}
            if(input .equals("command_kill")){action_event(input);}
            if(input .equals("help")){action_event(input);}
        }
    }

Here is the stack trace

Exception in thread "Thread-1" java.lang.NullPointerException
at Platypus_User$Inport.run(Platypus_User.java:64)

This is being ran in Eclipse on Mac OSX. A Null Pointer Exception occurs following the second catch block where the string is compared to "w" then if it is "w" the action_event method is called.

I have no clue why this would be happening. I would appreciate any advice.

Also, some advice: when you need to write something like this: if(input .equals("w")), it might be better to write it as: if ("w".equals(input)) since it will never be possible for this version to throw a NPE. — Stultuske
– Stultuske, Commented Apr 21, 2015 at 16:11
@Stultuske better use Set#contains() though such code repetitions make code look dirty — Alex Salauyou
– Alex Salauyou, Commented Apr 21, 2015 at 16:30

HappyCactus · Accepted Answer · 2015-04-21 16:12:12Z

3

I guess in is a BufferedReader. readLine will return null if End Of The Stream is reached.

See BufferedReader documentation

answered Apr 21, 2015 at 16:12

HappyCactus

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2 Comments

Nicholas McCann Over a year ago

Yes "in" is a Buffered Reader. How should I read user input then ?

HappyCactus Over a year ago

You should just check if input == null, and skip the checks

Ryan · Accepted Answer · 2015-04-21 16:12:52Z

1

BufferedReader.readLine() can return null, so check for null on input.

answered Apr 21, 2015 at 16:12

Ryan

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Comments

Alex Salauyou · Accepted Answer · 2015-04-21 16:37:53Z

First, avoid repetitions in code. You can collect all allowed inputs in one Set and then just check if it contains particular value--thus making code look much more readable, short and clean.

Second, you need to perform null check, because in.read() can return null as mentioned in other answers. null input could also be used for termination of while loop.

So I would rewrite your code as following:

Set<String> allowedInputs 
    = new HashSet<>(Arrays.asList("w", "a", "s", "d", "eat")); // <- add remaining allowed inputs here

String input = "";

while (input != null) {    
    try {
        input = in.readLine();
    } catch (IOException e1) {
        System.out.println(e1 + "Exception occured when reading user input");
    }
    try {
        Thread.sleep(USER_THROTTLE);
    } catch(Exception e) {
        System.out.println(toString() + " has input interrupted.");
    }
    if (input != null && allowedInputs.contains(input)) { // <- check if input is allowed
         action_event(input);
    }
}

Samuel Sunny Wilson · Accepted Answer · 2015-10-10 07:52:26Z

0

Never invoke methods on objects which are not initialized. input may hold a null value. So before comparing you must make sure that object is not null. Best approach to handle such situations would be by comparing a constant value with "input" instead of comparing input with a constant value. eg "w".equals(input)

answered Oct 10, 2015 at 7:52

Samuel Sunny Wilson

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