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- Question
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
- Correct Answer
- dokey
ExplanationThe & 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".
Operators and Assignments problems
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- 1. What will be the output of the program?
class Test { static int s; public static void main(String [] args) { Test p = new Test(); p.start(); System.out.println(s); } void start() { int x = 7; twice(x); System.out.print(x + " "); } void twice(int x) { x = x*2; s = x; } }
Options- A. 7 7
- B. 7 14
- C. 14 0
- D. 14 14 Discuss
Correct Answer: 7 14
Explanation:
The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.- 2. What will be the output of the program?
class BitShift { public static void main(String [] args) { int x = 0x80000000; System.out.print(x + " and "); x = x >>> 31; System.out.println(x); } }
Options- A. -2147483648 and 1
- B. 0x80000000 and 0x00000001
- C. -2147483648 and -1
- D. 1 and -2147483648 Discuss
Correct Answer: -2147483648 and 1
Explanation:
Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
Option B is incorrect because the output method print() always displays integers in base 10.
Option D is incorrect because this is the reverse order of the two output numbers.
- 3. What will be the output of the program?
class Test { public static void main(String [] args) { Test p = new Test(); p.start(); } void start() { boolean b1 = false; boolean b2 = fix(b1); System.out.println(b1 + " " + b2); } boolean fix(boolean b1) { b1 = true; return b1; } }
Options- A. true true
- B. false true
- C. true false
- D. false false Discuss
Correct Answer: false true
Explanation:
The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.- 4. What will be the output of the program?
class Test { public static void main(String [] args) { int x=20; String sup = (x < 15)? "small" : (x < 22)? "tiny" : "huge"; System.out.println(sup); } }
Options- A. small
- B. tiny
- C. huge
- D. Compilation fails Discuss
Correct Answer: tiny
Explanation:
This is an example of a nested ternary operator. The second evaluation (x < 22) is true, so the "tiny" value is assigned to sup.- 5. What will be the output of the program?
class Equals { public static void main(String [] args) { int x = 100; double y = 100.1; boolean b = (x = y); /* Line 7 */ System.out.println(b); } }
Options- A. true
- B. false
- C. Compilation fails
- D. An exception is thrown at runtime Discuss
Correct Answer: Compilation fails
Explanation:
The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables.Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.
- 6. 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.- 7. 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.- 8. 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.- 9. 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.
- 10. 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".
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- 1. What will be the output of the program?