return statement effect in Exception class

[Harikrishna Gorrepati]

Hi,

I have given the questions in the comments at Line # 10 and 11. Please advice

public class P199_Exception14_return_Ask {
    
    public static void main(String[] args) {
    	try {
    		printAverage(100, 0);
    	}
    	
    	catch(ArithmeticException e) {
    		System.out.println("ArithmeticException");
    		return;			//Q1. If I keep this return, After finally block code i.e., "After finally" will not be printed. Why ??
    					//Q2. Also, if return is there in both try and catch, Unreachable error code @ System.out.println("After finally");
    	}
    	
    	finally {
    		System.out.println("finally");
    	}
    	
    	System.out.println("After finally");
    }

    public static void printAverage(int totalSum, int totalNumber) throws ArithmeticException {
       return totalSum/totalNumber;  
    }
}

[Kevin Workman]

Q1- Why would it? The finally block has nothing to do with this question, by the way.
Q2- It's telling you exactly why. If you return from both a try and catch block, how will you ever reach the code after the try/catch?

[Harikrishna Gorrepati]

Kevin Workman wrote:Q1- Why would it? The finally block has nothing to do with this question, by the way.
Q2- It's telling you exactly why. If you return from both a try and catch block, how will you ever reach the code after the try/catch?

Hi Kevin,I changed the code and trying to understand the concept behind for these 2 points. Please advice

public class P199_Exception14_return_Ask {
    
    public static void main(String[] args) {
    	try {
    		printAverage(100, 0);
            //return;
    	}
    	
    	catch(ArithmeticException e) {
    		System.out.println("ArithmeticException");
    		return;			//Q1. If I keep this return, "Hello World" will not be printed. Why ??
    					//Q2. Also, if return is there in both try and catch, Unreachable error code @ System.out.println("After finally");
    	}
    	System.out.println("Hello World");
    }

    public static void printAverage(int totalSum, int totalNumber) throws ArithmeticException {
       return totalSum/totalNumber;  
    }
}

[Kevin Workman]

My answers do not change.

If you can't do it "by hand", then you should step through your program with a debugger to understand what it's doing.

[Matthew Brown]

In the following case:

public static void main(String[] args) {  
    int i = 10;
    if (i < 20) {
        return;
    }
    System.out.println("Hello World"); 
}

Would you expect "Hello World" to be printed? No - the method returns before you get there. It's the same in the example you've given. return exits the method, not the block it's in.

[Rohit Ramachandran]

Firstly, the code won't compile. You can't return a value from a void method.

Secondly, dude if you put a return in a try and a catch, how will you ever reach "After finally"?

And, I still don't understand why "After finally" is not being printed. It should, because once the exception is caught and dealt with, shouldn't it move on to the next line of code?

[Rohit Ramachandran]

Mathew Brown, the compiler doesn't know that i is actually < 20. But the compiler knows in the previous example that the code below won't ever be reached.

[Stephan van Hulst]

Rohit Ramachandran wrote:And, I still don't understand why "After finally" is not being printed. It should, because once the exception is caught and dealt with, shouldn't it move on to the next line of code?

Yes, the exception is caught and dealt with, but note the return statement in the catch block. Meaning that after the exception is dealt with (and the finally block is executed), the method returns.

[Rohit Ramachandran]

Hmm thank you.

[Harikrishna Gorrepati]

Hi Mat, Let me make it simple. Why line 10 is giving an error as given below

public class P199_Exception14_3 {
    public static void main(String[] args) {
	try {
	    System.out.println("try");
	    return;
	} catch(Exception e) {
	    System.out.println("Exception");
	    return;
	}
	 System.out.println("Hello World"); //Compiler Error : Unreachable code..Why ??
    }
}

[Rohit Ramachandran]

Dude, at line 3, the code can either reach the try part and then the catch part of the try/catch block. In both cases, you've put a return so there's no way you'll ever get to line 10. THe error says clearly, "Unreachable code"

[Harikrishna Gorrepati]

Even though I have only one return in catch block, Line#10 is still unreachable..Why..

public class P199_Exception14_3 {
    public static void main(String[] args) {
	try {
	    System.out.println("try");
	    throw new Exception("");
	} catch(Exception e) {
	    System.out.println("Exception");
	    return;
	}
	 System.out.println("Hello World"); //Unreachable code
    }
}

[Stephan van Hulst]

The compiler pretty much strips your catch block, and inlines it, because it sees that you will always throw an exception. It pretty much does this:

public class P199_Exception14_3 {
    public static void main(String[] args) {
      System.out.println("try");
      System.out.println("Exception");
      return;
     System.out.println("Hello World"); //Unreachable code
    }
}

So it sees that your last line will *never* be run.

[Rohit Ramachandran]

Because you have a "throw" in the try part of the try/catch block, which means each time the try block is entered, it ends up in the catch block. The compiler knows this.

Just modify the code to this-

public class P199_Exception14_3 {
    public static void main(String[] args) {
	try {
	    System.out.println("try");
	    int x=5/0;
	} catch(Exception e) {
	    System.out.println("Exception");
	    return;
	}
	 System.out.println("Hello World"); //Unreachable code
    }
}

Hello World is still unreachable but you won't get the unreachable code error since the compiler doesn't know that x=5/0 will throw an ArithmeticException: divide by zero.

Do you understand it now?

[Harikrishna Gorrepati]

Great !! Thanks Stephan & Rohit. I understand it now..