How to throw an IllegalArgumentException when an array is null or empty?

Question

    /**
     * This constructor accepts an array of points as input. Copy the points into the array points[]. 
     * 
     * @param  pts  input array of points 
     * @throws IllegalArgumentException if pts == null or pts.length == 0.
     */
    protected AbstractSorter(Point[] pts) throws IllegalArgumentException
    {
        try{
            for(int i = 0; i < pts.length; i++)
            {
                points[i] = pts[i];
            }
        }
        catch()
        {

        }
    }

I know this should be really simple, but how would I throw this exception with those conditions?

If `pts` is null you should probably throw `NullPointerException`. — markspace, Mar 07 '20 at 18:56
@kaya3 Yeah it appears I am very confused... I am new to java. So what would it look like then? Sorry — Andrew Wilken, Mar 07 '20 at 18:58
Generally this is done at the beginning of the method. Uses the throw statement. — NomadMaker, Mar 07 '20 at 18:59

score 2 · Answer 1 · edited Mar 07 '20 at 23:22

2

protected AbstractSorter(Point[] pts) throws IllegalArgumentException
{
        if(pts == null || pts.length ==0 )
          throw new IllegalArgumentException();
        for(int i = 0; i < pts.length; i++)
        {
            points[i] = pts[i];
        }
}

edited Mar 07 '20 at 23:22

nylanderDev

531
4
14

answered Mar 07 '20 at 18:59

Dinesh

1,046
1
8
17

1

This isn't valid Java code because the `catch` block doesn't declare a variable for the exception to be caught. But more importantly there shouldn't be a `try` or `catch` block at all. – kaya3 Mar 07 '20 at 19:00

Anurodh Singh · Accepted Answer · 2020-03-08T10:20:57.917

1

Your code should be like this,

protected AbstractSorter(Point[] pts) throws IllegalArgumentException
{
    if(pts == null || pts.length == 0 ){
      throw new IllegalArgumentException();
    }
    for(int i = 0; i < pts.length; i++)
    {
        points[i] = pts[i];
    }
}

edited Mar 08 '20 at 10:20

answered Mar 07 '20 at 19:09

Anurodh Singh

814
5
9

1

Where would an IllegalArgumentException be thrown in the try block ? – Arthur Attout Mar 07 '20 at 23:26
The try-catch block is for the other exceptions apart from IllegalArgumentException. – Anurodh Singh Mar 08 '20 at 04:40
Where would an Exception be thrown in the try block ? Why would you catch it ? – Arthur Attout Mar 08 '20 at 09:55
Yes, you correct. Now I have looked at the code I can there is no possible way for an exception in the try block. – Anurodh Singh Mar 08 '20 at 10:18

score 0 · Answer 3 · answered Mar 07 '20 at 19:05

0

/**
     * This constructor accepts an array of points as input. Copy the points into the array points[]. 
     * 
     * @param  pts  input array of points 
     * @throws IllegalArgumentException if pts == null or pts.length == 0.
     */
    protected AbstractSorter(Point[] pts) throws IllegalArgumentException
    {
        if(pts == null || pts.length == 0)
            throw new IllegalArgumentException();
        else{
            for(int i = 0; i < pts.length; i++)
                points[i] = pts[i];
        }
    }

This is what I have come up with, with your advice.

answered Mar 07 '20 at 19:05

Andrew Wilken

55
7

Nice one, yes, this is a correct solution. One thing I'll suggest is always use braces `{` ... `}` even for the `if` and `for` blocks here where there is only one statement inside them; it makes your code more readable, and makes it easier to edit without introducing bugs. – kaya3 Mar 07 '20 at 19:06
"else" is unnecessary after throw. Also it is unusual to answer your own question if there is a similar answer by someone else, and duplicate answers have little value. – tkruse Mar 08 '20 at 02:03

How to throw an IllegalArgumentException when an array is null or empty?

3 Answers3