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- Question
What will be the output of the program?
class Happy extends Thread { final StringBuffer sb1 = new StringBuffer(); final StringBuffer sb2 = new StringBuffer(); public static void main(String args[]) { final Happy h = new Happy(); new Thread() { public void run() { synchronized(this) { h.sb1.append("A"); h.sb2.append("B"); System.out.println(h.sb1); System.out.println(h.sb2); } } }.start(); new Thread() { public void run() { synchronized(this) { h.sb1.append("D"); h.sb2.append("C"); System.out.println(h.sb2); System.out.println(h.sb1); } } }.start(); } }
Options- A. ABBCAD
- B. ABCBCAD
- C. CDADACB
- D. Output determined by the underlying platform.
- Correct Answer
- Output determined by the underlying platform.
ExplanationCan you guarantee the order in which threads are going to run? No you can't. So how do you know what the output will be? The output cannot be determined.
More questions
- 1. What will be the output of the program?
int i = (int) Math.random();
Options- A. i = 0
- B. i = 1
- C. value of i is undetermined
- D. Statement causes a compile error Discuss
Correct Answer: i = 0
Explanation:
Math.random() returns a double value greater than or equal to 0 and less than 1. Its value is stored to an int but as this is a narrowing conversion, a cast is needed to tell the compiler that you are aware that there may be a loss of precision.The value after the decimal point is lost when you cast a double to int and you are left with 0.
- 2. What will be the output of the program?
class Tree { } class Pine extends Tree { } class Oak extends Tree { } public class Forest1 { public static void main (String [] args) { Tree tree = new Pine(); if( tree instanceof Pine ) System.out.println ("Pine"); else if( tree instanceof Tree ) System.out.println ("Tree"); else if( tree instanceof Oak ) System.out.println ( "Oak" ); else System.out.println ("Oops "); } }
Options- A. Pine
- B. Tree
- C. Forest
- D. Oops Discuss
Correct Answer: Pine
Explanation:
The program prints "Pine".- 3. What will be the output of the program?
public class StringRef { public static void main(String [] args) { String s1 = "abc"; String s2 = "def"; String s3 = s2; /* Line 7 */ s2 = "ghi"; System.out.println(s1 + s2 + s3); } }
Options- A. abcdefghi
- B. abcdefdef
- C. abcghidef
- D. abcghighi Discuss
Correct Answer: abcghidef
Explanation:
After line 7 executes, both s2 and s3 refer to a String object that contains the value "def". When line 8 executes, a new String object is created with the value "ghi", to which s2 refers. The reference variable s3 still refers to the (immutable) String object with the value "def".- 4. What will be the output of the program?
public class ObjComp { public static void main(String [] args ) { int result = 0; ObjComp oc = new ObjComp(); Object o = oc; if (o == oc) result = 1; if (o != oc) result = result + 10; if (o.equals(oc) ) result = result + 100; if (oc.equals(o) ) result = result + 1000; System.out.println("result = " + result); } }
Options- A. 1
- B. 10
- C. 101
- D. 1101 Discuss
Correct Answer: 1101
Explanation:
Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.- 5. What will be the output of the program?
String d = "bookkeeper"; d.substring(1,7); d = "w" + d; d.append("woo"); /* Line 4 */ System.out.println(d);
Options- A. wookkeewoo
- B. wbookkeeper
- C. wbookkeewoo
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
In line 4 the code calls a StringBuffer method, append() on a String object.- 6. What will be the output of the program?
interface Count { short counter = 0; void countUp(); } public class TestCount implements Count { public static void main(String [] args) { TestCount t = new TestCount(); t.countUp(); } public void countUp() { for (int x = 6; x>counter; x--, ++counter) /* Line 14 */ { System.out.print(" " + counter); } } }
Options- A. 0 1 2
- B. 1 2 3
- C. 0 1 2 3
- D. 1 2 3 4
- E. Compilation fails Discuss
Correct Answer: Compilation fails
Explanation:
The code will not compile because the variable counter is an interface variable that is by default final static. The compiler will complain at line 14 when the code attempts to increment counter.- 7. Which three form part of correct array declarations?
- public int a [ ]
- static int [ ] a
- public [ ] int a
- private int a [3]
- private int [3] a [ ]
- public final int [ ] a
Options- A. 1, 3, 4
- B. 2, 4, 5
- C. 1, 2, 6
- D. 2, 5, 6 Discuss
Correct Answer: 1, 2, 6
Explanation:
(1), (2) and (6) are valid array declarations.Option (3) is not a correct array declaration. The compiler complains with: illegal start of type. The brackets are in the wrong place. The following would work: public int[ ] a
Option (4) is not a correct array declaration. The compiler complains with: ']' expected. A closing bracket is expected in place of the 3. The following works: private int a []
Option (5) is not a correct array declaration. The compiler complains with 2 errors:
']' expected. A closing bracket is expected in place of the 3 and
<identifier> expected A variable name is expected after a[ ] .
- 8. 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.- 9. Which of the following line of code is suitable to start a thread?
class X implements Runnable { public static void main(String args[]) { /* Missing code? */ } public void run() {} }
Options- A. Thread t = new Thread(X);
- B. Thread t = new Thread(X); t.start();
- C. X run = new X(); Thread t = new Thread(run); t.start();
- D. Thread t = new Thread(); x.run(); Discuss
Correct Answer: X run = new X(); Thread t = new Thread(run); t.start();
Explanation:
Option C is suitable to start a thread.- 10. What will be the output of the program?
try { Float f1 = new Float("3.0"); int x = f1.intValue(); byte b = f1.byteValue(); double d = f1.doubleValue(); System.out.println(x + b + d); } catch (NumberFormatException e) /* Line 9 */ { System.out.println("bad number"); /* Line 11 */ }
Options- A. 9.0
- B. bad number
- C. Compilation fails on line 9.
- D. Compilation fails on line 11. Discuss
Correct Answer: 9.0
Explanation:
The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.
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- 1. What will be the output of the program?