The local variable might not have been initialized - Detect unchecked exception throw within a method

Question

I have some code with this structure:

public void method() { Object o; try { o = new Object(); } catch (Exception e) { //Processing, several lines throw new Error(); //Our own unchecked exception } doSomething(o); }

I have quite a few methods in which I have the same code in the catch block, so I want to extract it to a method so that I can save some lines. My problem is, that if I do that, I get a compiler error " The local variable o might not have been initialized".

public void method() { Object o; try { o = new Object(); } catch (Exception e) { handleError(); } //doSomething(o); compiler error } private void handleError() throws Error { //Processing, several lines throw new Error(); }

Is there any workaround?

Note:Instance variable not need to initialize but the local variable(variable inside a method) need to initialize — gjman2
– gjman2, Commented Aug 7, 2013 at 9:25

Sanjaya Liyanage · Accepted Answer · 2013-08-07 09:15:25Z

8

You need to initialize local variables before they are used as below

public void method() { Object o=null; try { o = new Object(); } catch (Exception e) { handleError(); } doSomething(o); }

You will not get the compilation failure until you use local variable which was not initialized

answered Aug 7, 2013 at 9:15

Sanjaya Liyanage

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2 Comments

Eusebius Over a year ago

Not really suitable if the variable has to be final for some reason. In this case I would favor the solution provided by @gyro-gearless.

user3310334 Over a year ago

This feels so gross. Is there no better style?

sanbhat · Accepted Answer · 2013-08-07 09:20:05Z

Since o is getting initialized within the try block and initializing o might throw an exception, java thinks that doSomething(o) statement might reach without o being initialized. So java wants o to be initialized incase new Object() throws exception.

So initializing o with null will fix the issue

public void method() { Object o = null; try { o = new Object(); //--> If new Object() throws exception then o remains uninitialized } catch (Exception e) { handleError(); } if(o != null) doSomething(o); }

I'd consider this an anti-pattern, as it clutters the code with unnecessary if-not-null checks! The try/catch mechanism is there to avoid such pattern: if the creation of some required objects fails, the function should leave through an exception.
True, i agree. My intention was to give the solution and explain why we get error

Juvanis · Accepted Answer · 2013-08-07 09:15:21Z

Initialize your object: Object o = null;, however watch out for the NullPointerExceptions that might be thrown when you give it to the method calls.

Gyro Gearless · Accepted Answer · 2013-08-07 09:23:03Z

Just put the doSomething(o) inside the try { } block:

public void method() { Object o; try { o = new Object(); doSomething(o); } catch (Exception e) { handleError(); } }

You perhaps dont want to execute doSomething() if the creation of your Object fails!

Mohsin Shaikh · Accepted Answer · 2013-08-07 09:37:25Z

Instance variable is the Object type so you should initialize value "null"

public void method() { Object o=null; try { o = new Object(); } catch (Exception e) { handleError(); } doSomething(o); }

Michał Kolenda · Accepted Answer · 2021-09-30 18:16:06Z

It happens because compiler doesn't know if handlerError always throws the Error. For example, there can be a logic in handleError method that makes it returns before throwing exception.

If you want to extract error handling into method, you can do this in this way:

public void method() { Object o; try { o = new Object(); } catch (Exception e) { throw handleError(); } System.out.println(o); } private Error handleError() { //Processing, several lines return new Error(); }

Now the compiler sees the throw word in catch block, and it is sure that the exception will be thrown.

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