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- Question
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 */ } }
Options- 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.
- Correct Answer
- C is printed before exiting with an error message.
ExplanationError 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).
Exceptions problems
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- 1. 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 "); } }
Options- A. hello throwit caught
- B. Compilation fails
- C. hello throwit RuntimeException caught after
- D. hello throwit caught finally after Discuss
Correct Answer: hello throwit caught finally after
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.
- 2. 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() {} }
Options- A. AC
- B. BC
- C. ACD
- D. ABCD Discuss
Correct Answer: ACD
Explanation:
There is no exception thrown, so all the code with the exception of the catch statement block is run.- 3. What will be the output of the program?
class PassA { public static void main(String [] args) { PassA p = new PassA(); p.start(); } void start() { long [] a1 = {3,4,5}; long [] a2 = fix(a1); System.out.print(a1[0] + a1[1] + a1[2] + " "); System.out.println(a2[0] + a2[1] + a2[2]); } long [] fix(long [] a3) { a3[1] = 7; return a3; } }
Options- A. 12 15
- B. 15 15
- C. 3 4 5 3 7 5
- D. 3 7 5 3 7 5 Discuss
Correct Answer: 15 15
Explanation:
Output: 15 15The reference variables a1 and a3 refer to the same long array object. When the [1] element is updated in the fix() method, it is updating the array referred to by a1. The reference variable a2 refers to the same array object.
So Output: 3+7+5+" "3+7+5
Output: 15 15 Because Numeric values will be added
- 4. What will be the output of the program?
class PassS { public static void main(String [] args) { PassS p = new PassS(); p.start(); } void start() { String s1 = "slip"; String s2 = fix(s1); System.out.println(s1 + " " + s2); } String fix(String s1) { s1 = s1 + "stream"; System.out.print(s1 + " "); return "stream"; } }
Options- A. slip stream
- B. slipstream stream
- C. stream slip stream
- D. slipstream slip stream Discuss
Correct Answer: slipstream slip stream
Explanation:
When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".- 5. What will be the output of the program?
public class Test { public static void leftshift(int i, int j) { i <<= j; } public static void main(String args[]) { int i = 4, j = 2; leftshift(i, j); System.out.printIn(i); } }
Options- A. 2
- B. 4
- C. 8
- D. 16 Discuss
Correct Answer: 4
Explanation:
Java only ever passes arguments to a method by value (i.e. a copy of the variable) and never by reference. Therefore the value of the variable i remains unchanged in the main method.If you are clever you will spot that 16 is 4 multiplied by 2 twice, (4 * 2 * 2) = 16. If you had 16 left shifted by three bits then 16 * 2 * 2 * 2 = 128. If you had 128 right shifted by 2 bits then 128 / 2 / 2 = 32. Keeping these points in mind, you don't have to go converting to binary to do the left and right bit shifts.
- 6. What will be the output of the program?
public class Foo { public static void main(String[] args) { try { return; } finally { System.out.println( "Finally" ); } } }
Options- A. Finally
- B. Compilation fails.
- C. The code runs with no output.
- D. An exception is thrown at runtime. 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: Finally
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:- 7. What will be the output of the program?
public class MyProgram { public static void main(String args[]) { try { System.out.print("Hello world "); } finally { System.out.println("Finally executing "); } } }
Options- A. Nothing. The program will not compile because no exceptions are specified.
- B. Nothing. The program will not compile because no catch clauses are specified.
- C. Hello world.
- D. Hello world Finally executing Discuss
Correct Answer: Hello world Finally executing
Explanation:
Finally clauses are always executed. The program will first execute the try block, printing Hello world, and will then execute the finally block, printing Finally executing.Option A, B, and C are incorrect based on the program logic described above. Remember that either a catch or a finally statement must follow a try. Since the finally is present, the catch is not required.
- 8. What will be the output of the program?
public class Test { public static void aMethod() throws Exception { try /* Line 5 */ { throw new Exception(); /* Line 7 */ } finally /* Line 9 */ { System.out.print("finally "); /* Line 11 */ } } public static void main(String args[]) { try { aMethod(); } catch (Exception e) /* Line 20 */ { System.out.print("exception "); } System.out.print("finished"); /* Line 24 */ } }
Options- A. finally
- B. exception finished
- C. finally exception finished
- D. Compilation fails Discuss
Correct Answer: finally exception finished
Explanation:
This is what happens:(1) The execution of the try block (line 5) completes abruptly because of the throw statement (line 7).
(2) The exception cannot be assigned to the parameter of any catch clause of the try statement therefore the finally block is executed (line 9) and "finally" is output (line 11).
(3) The finally block completes normally, and then the try statement completes abruptly because of the throw statement (line 7).
(4) The exception is propagated up the call stack and is caught by the catch in the main method (line 20). This prints "exception".
(5) Lastly program execution continues, because the exception has been caught, and "finished" is output (line 24).
- 9. 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(); } }
Options- A. BD
- B. BCD
- C. BDE
- D. BCDE Discuss
Correct Answer: BDE
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.- 10. 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");
Options- A. finished
- B. Exception
- C. Compilation fails.
- D. Arithmetic Exception Discuss
Correct Answer: Compilation fails.
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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- 1. What will be the output of the program?