Question

Question 5 (10 points) Assume you have the following code, with blanks to be filled in below. public static void g () ( throw new () public static void f() try ) catch ( 12), e) System.out.print ("A") ; return } catch ( System.out. print ("B") throw ei e) ( 13) catch ( 14) e) ( System.out.print ("C") ; } finally System.out.print ("D") ; System.out.print ("E") ; } public static void main (String (] args) try f() } catch ( System.out.print ("F") ; } finally System.out.print ("G") ; (5) e) System.out.print ("H") ; } Assume each of the following sets were inserted into the code above. For each set, determine if it would compile. If it would compile, write what the output would be be. 1) NullPointerException 1) ArithmeticException 2) RuntimeException 3) NullPointerException 3) RuntimeException 4) Exception 5) NullPointerException 5) Exception 1) Exception 2) ArithmeticException 2) ArithmeticException 3) RuntimeException 4) Exception 5) Exception 1) ArithmeticException 2) NullPointerException 3) Exception 4) Throwable 5) Exception 4) Exception Compiles: YN If Y, Output: Compiles: YN If Y, Output: Compiles: YN If Y, Output: ADa Compiles: If Y, Output: ABCF Ba

Answer 1

Solution:

Exceptions:

Exceptions are generally considered as the events that occur during program execution and normal flow of program execution alters.

There are various kinds of exceptions like RuntimeException, NullPointerException, ArithmeticException etc..

It is necessary to handle these exceptions when they occur in our execution. We catch them using catch() blocks present after try blocks on our code.

Given Four sets with various options to be filled in given code.

The outputs that are obtained after placing these different options in our code are as shown below.

First Set:

The program gives error.

Here when the exception is thrown, it must be handled in the same function when it is not declared.

So, this statement as shown in the below screenshot does not compile and gives you a error that it must be caught or declared to be thrown.

Main.java LTTPL SSPI 2 public static void g (){ throw new Exception(); 4 5 public static void f() try{ g(); } catch(ArithmeticException e){ System.out.print("A"); return; catch(RuntimeException e) System.out print("8"); return catch(Exception e) System.out.print("C"); 6 7 87 10 11T 12 13 14 15 16 return } finally System.out print ("D"); 17 18 19 System.out.print("E"); 20 21 22 public static void main(Stning args) try{ f ) catch(Exception e) System.out.print("F); } finally System.out.print("G) 23 24 25 26 27 T 28 29 30 System.out.print (H"); 31 Main.java:3: error: unreported exception Exception; aust be caught or declared to be thrown

Second Set:

The code compiles and produces output ADGH.

The Arithmetic Exception is caught when it is raised and the A is displayed on console. And the statements in the finally block are displayed.

Main.java 2 public static void g()( throw new ArithmeticException (); 3 4 5public static void f() try{ g(); catch(ArithmeticException e){ System.out.print("A"); 6 7 9 return ; 10 }catch(Exception e){ System.out.print("B"); return ; 11 12 13 catch(Throwable e){ System.out.print("C"); return; 14 15 16 } finally( System.out.print("D"); 17 18 19 System.out print("E"); 20 21 22 public static void main(String[] args){ tryf f) ) catch(Exception e){ System.out.print("F"); ) finally{ system.out.print("G"); 23 24 25 T 26 27 28 29 System.out.print("H); 30 31 I ADGH >

Third Set:

The code gives error.

In the main function when f() is called, it calls g() in turn and Null Pointer Exception is handled.

But in the main method again Null pointer Exception is being caught after try statement.

This raises error that NullPointerException has already been caught.

Main.java UTPW SSP 2 public st atic void g()E throw new NullPointerException(); } 3 4 5 public static void f() try{ g(); } catch(RuntimeException e){ System.out.print("A"); return ; catch(Nul1PointerException e)( System.out.print("B"); return ; 8 10 11 12 13 14 } catch(Exception e){ System.out.print("C"); return ; 15 16 17 } finally{ System.out.print ("D"); 18 19 System.out.print("E"); 20 21 22 public static void main(String[] args){ 23T try{ fO; catch(NullPointerException e){ system.out.print("F"); } finally{ System.out.print("G"); 24 25 26 27 28 29 System.out.print("H"); }} Compilation failed due to following erroris Main.java:11: error: exception NullPointer Exception has already been caught 30 31 n67 M

Fourth set:

The code compiles and produces output as BDGH.

The Arithmetic exception is thrown and it is subset of RuntimeException. So it is caught by it and prints B on console.

Then the statements that are present in the finally block are executed.

Main.java nrew sser 2 public static void g ()f throw new ArithmeticException(); 3 4 5 public static void f (O try{ g(); } catch(Nul1PointerException e){ System.out.print ("A"); return ; } catch (RuntimeException e){ System.out.print("B"); return ; 6 7 8 10 11- 12 13 catch(Exception e){ System.out.print("C"); 14 15 return 16 } finally{ System.out.print("D"); 17- 18 19 System.out.print("E"); 20 21 22 public static void main(String [] args){ try{ f(); } catch(Exception e){ System.out.print("F"); finally{ System.out.print ("G"); System.out.print("H"); input BDGH 222 222 2 m m