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Java Programming - Exceptions

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Home » Engineering » Java Programming » Exceptions » Finding the output

Exercise :: Exceptions - Finding the output

1. 

What will be the output of the program? 

public class Foo 
{  
    public static void main(String[] args) 
    {
        try 
        { 
            return; 
        } 
        finally 
        {
            System.out.println( "Finally" ); 
        } 
    } 
}

A. Finally
B. Compilation fails.
C. The code runs with no output.
D. An exception is thrown at runtime.

Answer: Option A

Explanation:

If you put a finally block after a try and its associated catch blocks, then once execution enters the try block, the code in that finally block will definitely be executed except in the following circumstances:
  1. An exception arising in the finally block itself.
  2. The death of the thread.
  3. The use of System.exit()
  4. Turning off the power to the CPU.

I suppose the last three could be classified as VM shutdown.

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2. 

What will be the output of the program? 

try 
{ 
    int x = 0; 
    int y = 5 / x; 
} 
catch (Exception e) 
{
    System.out.println("Exception"); 
} 
catch (ArithmeticException ae) 
{
    System.out.println(" Arithmetic Exception"); 
} 
System.out.println("finished");

A. finished
B. Exception
C. Compilation fails.
D. Arithmetic Exception

Answer: Option C

Explanation:

Compilation fails because ArithmeticException has already been caught. ArithmeticException is a subclass of java.lang.Exception, by time the ArithmeticException has been specified it has already been caught by the Exception class.

If ArithmeticException appears before Exception, then the file will compile. When catching exceptions the more specific exceptions must be listed before the more general (the subclasses must be caught before the superclasses).

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3. 

What will be the output of the program? 

public class X 
{  
    public static void main(String [] args) 
    {
        try 
        {
            badMethod();  
            System.out.print("A"); 
        }  
        catch (Exception ex) 
        {
            System.out.print("B");  
        } 
        finally 
        {
            System.out.print("C"); 
        } 
        System.out.print("D"); 
    }  
    public static void badMethod() 
    {
        throw new Error(); /* Line 22 */
    } 
}

A. ABCD
B. Compilation fails.
C. C is printed before exiting with an error message.
D. BC is printed before exiting with an error message.

Answer: Option C

Explanation:

Error is thrown but not recognised line(22) because the only catch attempts to catch an Exception and Exception is not a superclass of Error. Therefore only the code in the finally statement can be run before exiting with a runtime error (Exception in thread "main" java.lang.Error).

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4. 

What will be the output of the program? 

public class X 
{  
    public static void main(String [] args) 
    {
        try 
        {
            badMethod();  
            System.out.print("A");  
        } 
        catch (RuntimeException ex) /* Line 10 */
        { 
            System.out.print("B"); 
        } 
        catch (Exception ex1) 
        { 
            System.out.print("C"); 
        } 
        finally 
        {
            System.out.print("D"); 
        } 
        System.out.print("E"); 
    } 
    public static void badMethod() 
    { 
        throw new RuntimeException(); 
    } 
}

A. BD
B. BCD
C. BDE
D. BCDE

Answer: Option C

Explanation:

A Run time exception is thrown and caught in the catch statement on line 10. All the code after the finally statement is run because the exception has been caught.

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5. 

What will be the output of the program? 

public class RTExcept 
{
    public static void throwit () 
    {
        System.out.print("throwit ");
        throw new RuntimeException();
    }
    public static void main(String [] args) 
    {
        try 
        {
            System.out.print("hello ");
            throwit();
        }
        catch (Exception re ) 
        {
            System.out.print("caught ");
        }
        finally 
        {
            System.out.print("finally ");
        }
        System.out.println("after ");
    }
}

A. hello throwit caught
B. Compilation fails
C. hello throwit RuntimeException caught after
D. hello throwit caught finally after

Answer: Option D

Explanation:

The main() method properly catches and handles the RuntimeException in the catch block, finally runs (as it always does), and then the code returns to normal.

A, B and C are incorrect based on the program logic described above. Remember that properly handled exceptions do not cause the program to stop executing.

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