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- Question
Which four can be thrown using the throw statement?
- Error
- Event
- Object
- Throwable
- Exception
- RuntimeException
Options- A. 1, 2, 3 and 4
- B. 2, 3, 4 and 5
- C. 1, 4, 5 and 6
- D. 2, 4, 5 and 6
- Correct Answer
- 1, 4, 5 and 6
ExplanationThe (1), (4), (5) and (6) are the only four that can be thrown.An Error is a subclass of Throwable that indicates serious problems that a reasonable application should not try to catch.
The Throwable class is the superclass of all errors and exceptions in the Java language.
The class Exception and its subclasses are a form of Throwable that indicates conditions that a reasonable application might want to catch (checked exceptions)
RuntimeException is the superclass of those exceptions that can be thrown during the normal operation of the Java Virtual Machine.
Exceptions problems
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- 1. At Point X on line 5, which code is necessary to make the code compile?
public class ExceptionTest { class TestException extends Exception {} public void runTest() throws TestException {} public void test() /* Point X */ { runTest(); } }
Options- A. No code is necessary.
- B. throws Exception
- C. catch ( Exception e )
- D. throws RuntimeException Discuss
Correct Answer: throws Exception
Explanation:
Option B is correct. This works because it DOES throw an exception if an error occurs.Option A is wrong. If you compile the code as given the compiler will complain:
"unreported exception must be caught or declared to be thrown" The class extends Exception so we are forced to test for exceptions.
Option C is wrong. The catch statement belongs in a method body not a method specification.
Option D is wrong. TestException is a subclass of Exception therefore the test method, in this example, must throw TestException or some other class further up the Exception tree. Throwing RuntimeException is just not on as this belongs in the java.lang.RuntimeException branch (it is not a superclass of TestException). The compiler complains with the same error as in A above.
- 2. Which answer most closely indicates the behavior of the program?
public class MyProgram { public static void throwit() { throw new RuntimeException(); } public static void main(String args[]) { try { System.out.println("Hello world "); throwit(); System.out.println("Done with try block "); } finally { System.out.println("Finally executing "); } } }
Options- A. The program will not compile.
- B. The program will print Hello world, then will print that a RuntimeException has occurred, then will print Done with try block, and then will print Finally executing.
- C. The program will print Hello world, then will print that a RuntimeException has occurred, and then will print Finally executing.
- D. The program will print Hello world, then will print Finally executing, then will print that a RuntimeException has occurred. Discuss
Correct Answer: The program will print Hello world, then will print Finally executing, then will print that a RuntimeException has occurred.
Explanation:
Once the program throws a RuntimeException (in the throwit() method) that is not caught, the finally block will be executed and the program will be terminated. If a method does not handle an exception, the finally block is executed before the exception is propagated.- 3. Which statement is true?
Options- A. catch(X x) can catch subclasses of X where X is a subclass of Exception.
- B. The Error class is a RuntimeException.
- C. Any statement that can throw an Error must be enclosed in a try block.
- D. Any statement that can throw an Exception must be enclosed in a try block. Discuss
Correct Answer: catch(X x) can catch subclasses of X where X is a subclass of Exception.
Explanation:
Option A is correct. If the class specified in the catch clause does have subclasses, any exception object that subclasses the specified class will be caught as well.Option B is wrong. The error class is a subclass of Throwable and not Runtime Exception.
Option C is wrong. You do not catch this class of error.
Option D is wrong. An exception can be thrown to the next method higher up the call stack.
- 4. Which statement is true?
Options- A. A try statement must have at least one corresponding catch block.
- B. Multiple catch statements can catch the same class of exception more than once.
- C. An Error that might be thrown in a method must be declared as thrown by that method, or be handled within that method.
- D. Except in case of VM shutdown, if a try block starts to execute, a corresponding finally block will always start to execute. Discuss
- An exception arising in the finally block itself.
- The death of the thread.
- The use of System.exit()
- Turning off the power to the CPU.
Correct Answer: Except in case of VM shutdown, if a try block starts to execute, a corresponding finally block will always start to execute.
Explanation:
A is wrong. A try statement can exist without catch, but it must have a finally statement.B is wrong. A try statement executes a block. If a value is thrown and the try statement has one or more catch clauses that can catch it, then control will be transferred to the first such catch clause. If that catch block completes normally, then the try statement completes normally.
C is wrong. Exceptions of type Error and RuntimeException do not have to be caught, only checked exceptions (java.lang.Exception) have to be caught. However, speaking of Exceptions, Exceptions do not have to be handled in the same method as the throw statement. They can be passed to another method.
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:
- 5. Given that all methods of class FileOutputStream, including close(), throw an IOException, which of these is true?
import java.io.*; public class MyProgram { public static void main(String args[]) { FileOutputStream out = null; try { out = new FileOutputStream("test.txt"); out.write(122); } catch(IOException io) { System.out.println("IO Error."); } finally { out.close(); } } }
Options- A. This program will compile successfully.
- B. This program fails to compile due to an error at line 4.
- C. This program fails to compile due to an error at line 6.
- D. This program fails to compile due to an error at line 18. Discuss
Correct Answer: This program fails to compile due to an error at line 18.
Explanation:
Any method (in this case, the main() method) that throws a checked exception (in this case, out.close() ) must be called within a try clause, or the method must declare that it throws the exception. Either main() must declare that it throws an exception, or the call to out.close() in the finally block must fall inside a (in this case nested) try-catch block.- 6. What will be the output of the program?
class A { public A(int x){} } class B extends A { } public class test { public static void main (String args []) { A a = new B(); System.out.println("complete"); } }
Options- A. It compiles and runs printing nothing
- B. Compiles but fails at runtime
- C. Compile Error
- D. Prints "complete" Discuss
Correct Answer: Compile Error
Explanation:
No constructor has been defined for class B therefore it will make a call to the default constructor but since class B extends class A it will also call the Super() default constructor.Since a constructor has been defined in class A java will no longer supply a default constructor for class A therefore when class B calls class A's default constructor it will result in a compile error.
- 7. What will be the output of the program?
public class Test138 { public static void stringReplace (String text) { text = text.replace ('j' , 'c'); /* Line 5 */ } public static void bufferReplace (StringBuffer text) { text = text.append ("c"); /* Line 9 */ } public static void main (String args[]) { String textString = new String ("java"); StringBuffer textBuffer = new StringBuffer ("java"); /* Line 14 */ stringReplace(textString); bufferReplace(textBuffer); System.out.println (textString + textBuffer); } }
Options- A. java
- B. javac
- C. javajavac
- D. Compile error Discuss
Correct Answer: javajavac
Explanation:
A string is immutable, it cannot be changed, that's the reason for the StringBuffer class. The stringReplace method does not change the string declared on line 14, so this remains set to "java".Method parameters are always passed by value - a copy is passed into the method - if the copy changes, the original remains intact, line 5 changes the reference i.e. text points to a new String object, however this is lost when the method completes. The textBuffer is a StringBuffer so it can be changed.
This change is carried out on line 9, so "java" becomes "javac", the text reference on line 9 remains unchanged. This gives us the output of "javajavac"
- 8. What will be the output of the program?
public class Test178 { public static void main(String[] args) { String s = "foo"; Object o = (Object)s; if (s.equals(o)) { System.out.print("AAA"); } else { System.out.print("BBB"); } if (o.equals(s)) { System.out.print("CCC"); } else { System.out.print("DDD"); } } }
Options- A. AAACCC
- B. AAADDD
- C. BBBCCC
- D. BBBDDD Discuss
Correct Answer: AAACCC
- 9. What will be the output of the program?
String x = new String("xyz"); String y = "abc"; x = x + y;How many String objects have been created?
Options- A. 2
- B. 3
- C. 4
- D. 5 Discuss
Correct Answer: 4
Explanation:
Line 1 creates two, one referred to by x and the lost String "xyz". Line 2 creates one (for a total of three). Line 3 creates one more (for a total of four), the concatenated String referred to by x with a value of "xyzabc".- 10. What will be the output of the program?
public class ExamQuestion7 { static int j; static void methodA(int i) { boolean b; do { b = i<10 | methodB(4); /* Line 9 */ b = i<10 || methodB(8); /* Line 10 */ }while (!b); } static boolean methodB(int i) { j += i; return true; } public static void main(String[] args) { methodA(0); System.out.println( "j = " + j ); } }
Options- A. j = 0
- B. j = 4
- C. j = 8
- D. The code will run with no output Discuss
Correct Answer: j = 4
Explanation:
The lines to watch here are lines 9 & 10. Line 9 features the non-shortcut version of the OR operator so both of its operands will be evaluated and therefore methodB(4) is executed.However line 10 has the shortcut version of the OR operator and if the 1st of its operands evaluates to true (which in this case is true), then the 2nd operand isn't evaluated, so methodB(8) never gets called.
The loop is only executed once, b is initialized to false and is assigned true on line 9. Thus j = 4.
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- 1. At Point X on line 5, which code is necessary to make the code compile?