Why is this exception is not printed? Why is it showing an error?

Question

If I am trying to print the value of "a" why is it showing an error? Why has the exception become an error?

class Ankit1
{
    public static void main(String args[])
    {
        float d,a;
        try
        {
            d=0;
            a=44/d;
            System.out.print("It's not gonna print: "+a); // if exception doesn't occur then it will print and it will go on to the catch block
        }
        catch (ArithmeticException e)
        {
            System.out.println("a:" + a); // why is this an error??
        }
    }
}

A follow-up question here: stackoverflow.com/questions/12130800/… — vivek_jonam
– vivek_jonam, Commented Aug 26, 2012 at 14:24
The follow up is slightly different and slightly related, I'm kinda happy to allow both. — Kev
– Kev, Commented Aug 28, 2012 at 2:59

Bharat Sinha · Accepted Answer · 2012-08-26 14:07:35Z

6

If you see the error

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
    The local variable a may not have been initialized

    at your.package.Ankit1.main(Ankit1.java:18)

which clearly states The local variable a may not have been initialized

You're getting this error as your variable a wasn't initialized.

And if you want to print the error message try printing... e.getMessage() or p.printStackTrace() for complete stack trace.

To fix this simple initialize a with some value like this...

float a = 0;

edited Aug 26, 2012 at 14:07

answered Aug 26, 2012 at 13:53

Bharat Sinha

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2 Comments

Ankit Over a year ago

after initialization Undefined variable or class name: ae System.out.println("Print hoga"+ae.getMessage());

Bharat Sinha Over a year ago

I corrected that... try System.out.println("Print hoga"+e.getMessage()); but that too after you initialize a

Heisenbug · Accepted Answer · 2012-08-26 13:56:41Z

3

"if i am trying to print value of "a" why its showing error?

Because dividing by zero throws an exception before a is initialized.

For printing the error you can print exception message or the whole stacktrace:

catch(ArithmeticException e)
{
   System.out.println(e.getMessage());
   e.printStackTrace();
}

answered Aug 26, 2012 at 13:56

Heisenbug

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Comments

Community · Accepted Answer · 2017-05-23 11:45:26Z

3

a doesn't have any value. As exception happened in 44/d; statement as no value is in a probably.

Ankit1.java:14: variable a might not have been initialized
            System.out.println("Print hoga"+a);//why error come??

Its because the variable a is not initialized.

Also there won't be any ArithmeticException thrown for this 44/d statement because it has float operation so there is no Divide-by-zero Exception Instead Infinity will be the result.
For more see here

edited May 23, 2017 at 11:45

CommunityBot

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answered Aug 26, 2012 at 13:54

vivek_jonam

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1 Comment

vivek_jonam Over a year ago

@Ankit yes. Printing a variable that is not being initialized causes an error.

Ilya · Accepted Answer · 2012-08-26 13:52:18Z

2

a wasn't initialized
initialize default values for d and a

float d = 0.0f;  
float a = 0.0f;

or use Float instead of float

Float a = null;

answered Aug 26, 2012 at 13:52

Ilya

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1 Comment

Ankit Over a year ago

Undefined variable or class name: ae System.out.println("Print hoga"+ae.getMessage());//it is showing this error

Cratylus · Accepted Answer · 2012-08-26 14:08:29Z

1

You define float d,a; but you are not initialize them. If you don't also later, before you use them it is a compile time error.
In your try you do:
d=0; a=44/d;

But since you initialize them in a try and you access them inside the catch the compiler complains that a is not initialized. If you replaced with d you would also get the same error.
To solve this do:

float d = 0,a = 0;

Always initialize your local variables

answered Aug 26, 2012 at 14:08

Cratylus

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