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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
- Correct Answer
- slipstream slip stream
ExplanationWhen 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".
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- 1. 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.
- 2. What will be the output of the program?
class Test { public static void main(String [] args) { int x= 0; int y= 0; for (int z = 0; z < 5; z++) { if (( ++x > 2 ) || (++y > 2)) { x++; } } System.out.println(x + " " + y); } }
Options- A. 5 3
- B. 8 2
- C. 8 3
- D. 8 5 Discuss
Correct Answer: 8 2
Explanation:
The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.- 3. What will be the output of the program?
class Test { public static void main(String [] args) { int x= 0; int y= 0; for (int z = 0; z < 5; z++) { if (( ++x > 2 ) && (++y > 2)) { x++; } } System.out.println(x + " " + y); } }
Options- A. 5 2
- B. 5 3
- C. 6 3
- D. 6 4 Discuss
Correct Answer: 6 3
Explanation:
In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.- 4. What will be the output of the program?
class BoolArray { boolean [] b = new boolean[3]; int count = 0; void set(boolean [] x, int i) { x[i] = true; ++count; } public static void main(String [] args) { BoolArray ba = new BoolArray(); ba.set(ba.b, 0); ba.set(ba.b, 2); ba.test(); } void test() { if ( b[0] && b[1] | b[2] ) count++; if ( b[1] && b[(++count - 2)] ) count += 7; System.out.println("count = " + count); } }
Options- A. count = 0
- B. count = 2
- C. count = 3
- D. count = 4 Discuss
Correct Answer: count = 3
Explanation:
The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test.- 5. What will be the output of the program?
class SSBool { public static void main(String [] args) { boolean b1 = true; boolean b2 = false; boolean b3 = true; if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */ System.out.print("ok "); if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/ System.out.println("dokey"); } }
Options- A. ok
- B. dokey
- C. ok dokey
- D. No output is produced
- E. Compilation error Discuss
Correct Answer: dokey
Explanation:
The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".- 6. 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
- 7. 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.- 8. 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.
- 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 (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. Discuss
Correct Answer: C is printed before exiting with an error message.
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).- 10. 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:
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- 1. What will be the output of the program?