array variable might not have been initialized error

Question

I created class Supervisor which contains the constructor and a toString method. However when I try to print an Index of the array an error occur,"Variable svArray might not have been initialized. What can I do to solve this ?

Supervisor[] svArray;
if (mainChoice == 3){
        String ID, name, department, resarch;
        System.out.println("Enter How Many Supervisor you want to enter : ");
        svNumber = input.nextInt();
        svArray = new Supervisor[svNumber];
        for (int i = 0; i < svNumber; i++) {
            System.out.println("---Enter Supervisor " + (i + 1) + " Of " + svNumber + " ---");
            System.out.println("Enter staff ID : ");
            ID = input.next();
            System.out.println("Enter name : ");
            name = input.next();
            System.out.println("Enter department : ");
            department = input.next();
            svArray[i]=new Supervisor(ID, name, department);\
        }


    }
System.out.print(svArray[1].toString)
//Error, variable svArray might not have been initialized

You can't use a un-initialized variable in java. what about if mainChoice is not equal to 3. — Braj
– Braj, Commented Apr 14, 2014 at 16:06

Kuba Spatny · Accepted Answer · 2014-04-14 16:30:43Z

3

The problem is that on this line:

System.out.print(svArray[1].toString)

svArray hasn't been initialized in case that mainChoice != 3, because you initialize the array inside of if statement which is not executed.

To overcome this problem you can initialize the array to null:

Supervisor[] svArray = null;

And then check for null:

if(svArray != null) System.out.print(svArray[1].toString());

edited Apr 14, 2014 at 16:30

answered Apr 14, 2014 at 16:11

Kuba Spatny

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4 Comments

JonK Over a year ago

This doesn't compile because arrays don't have a toString attribute (admittedly the error is present in the OP's code as well).

Kuba Spatny Over a year ago

@JonK but you call toString on first element of said array which is Supervisor type

JonK Over a year ago

But you aren't calling toString(). That's my point. Method calls have brackets.

Kuba Spatny Over a year ago

@JonK Oh I see, thanks! I didn't noticed that and just blindly copy&pasted.

Elemental · Accepted Answer · 2014-04-14 16:09:57Z

2

Simply put you initialize the array inside the if statement but use it outside. So if the content of the if statement never runs (that is whenever mainChoice is not 3) the array is used at the end without being initialized.

answered Apr 14, 2014 at 16:09

Elemental

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Comments

matt · Accepted Answer · 2014-04-14 16:10:10Z

1

If mainChoice is not equal to 3, svArray never gets initialized. You need to either place the print statement into the if block or unconditionally initialize svArray[1] to some value.

answered Apr 14, 2014 at 16:10

matt

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Comments

JonK · Accepted Answer · 2014-04-14 16:25:19Z

Local variables differ from member variables in that they don't get automatically initialised to a default value for you. Java doesn't allow you to use a variable that hasn't been initialised yet, and the JVM can't know if mainChoice is always going to be 3. Because of this, the compiler can't be certain that your array will have been initialised by the time the call to System.out.print(svArray[1].toString) is attempted - hence your error.

It's not clear why you're declaring svArray outside of your loop, so this code may not be entirely appropriate:

if (mainChoice == 3){
    String ID, name, department, resarch;
    System.out.println("Enter How Many Supervisor you want to enter : ");
    svNumber = input.nextInt();
    Supervisor[] svArray = new Supervisor[svNumber];
    for (int i = 0; i < svNumber; i++) {
        System.out.println("---Enter Supervisor " + (i + 1) + " Of " 
            + svNumber + " ---");
        System.out.println("Enter staff ID : ");
        ID = input.next();
        System.out.println("Enter name : ");
        name = input.next();
        System.out.println("Enter department : ");
        department = input.next();
        svArray[i]=new Supervisor(ID, name, department);\
    }
    System.out.print(svArray[1].toString())
}

There are three important changes to note here.

The declaration and initialisation of svArray now occur on the same line within your if statement.
The System.out.println call now also happens inside your if statement so that svArray is still in scope.
It now correctly invokes the toString() method, rather than attempting to access an attribute called toString (which doesn't exist).

Elijah Dayan · Accepted Answer · 2016-04-23 20:48:02Z

0

 svArray[i]=new Supervisor(ID, name, department);

You initialized the above array inside the if statement. So when you're saying:System.out.print(svArray[1].toString)Java does not know that you initialized it inside the if statement. anything inside a block{} has a scope that is only valid inside that specific block segment, if that block is not the class type. Suppose you define something out of the class block then you will not get your variable or declaration working within the class block. However, if you use ArrayList then you may not need to initialize the size of it so just simply declaring the ArrayList in the class block will let you use it/initialize it in other methods/ statements.

answered Apr 23, 2016 at 20:48

Elijah Dayan

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