using method that take variable number of parameters

Question

I am trying to make a method that build new object of the class (PhoneBook) using different constructors according to the number of parameters ,, but it gives an error

( Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0 )

public static PhoneBook createObjects(String fName,String lName,String num,String...optional)
{
    n++;    
    if (optional[0]==null)
        ArrayOfObjects[n]=new PhoneBook(fName,lName,num);
    else if (optional[1]==null)
        ArrayOfObjects[n]=new PhoneBook(fName,lName,num,optional[0]);

    return ArrayOfObjects[n];   
}

@Vipar maybe a PhoneBook array declared somewhere else in OP's code (not relevant to the question)? — Luiggi Mendoza
– Luiggi Mendoza, Commented Dec 9, 2012 at 19:04
@Vipar actually I start n with 0 , but why this array never take values ? — Mohamed Mosaad
– Mohamed Mosaad, Commented Dec 9, 2012 at 19:14

NPE · Accepted Answer · 2012-12-09 19:02:50Z

3

Instead of checking whether optional[0] == null, you should examine optional.length to determine if the optional parameter is present.

The same goes for optional[1].

answered Dec 9, 2012 at 19:02

NPE

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Comments

mschultz · Accepted Answer · 2012-12-09 19:09:16Z

0

Rather than checking optional[0] and optional[1] here, you should check to optional.length. Also, keep in mind that optional itself may well be null, so something like:

if(optional != null) {
    if(optional.length > 0) {
        // I now know that optional has at least one element in it, and optional[0] should be valid, though I don't know that it is non-null.
        if(optional.length > 1) {
            // I now know that optional[1] is valid, though I do not know it is non-null.
        }
    }
}

if you NEED non-null:

if(optional.length > 0 && optional[0] != null)

The second part, optional[0] != null will only be called if the first evaluates to true.

answered Dec 9, 2012 at 19:09

mschultz

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