Operators and Assignments - Finding the output (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);
}
}

A.	5 2
B.	5 3
C.	6 3
D.	6 4

Answer: Option C

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.

10.

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);
}
}

A.	small
B.	tiny
C.	huge
D.	Compilation fails

Answer: Option B

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.

11.

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);
}
}

A.	true
B.	false
C.	Compilation fails
D.	An exception is thrown at runtime

Answer: Option C

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.

12.

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);
}
}

A.	-2147483648 and 1
B.	0x80000000 and 0x00000001
C.	-2147483648 and -1
D.	1 and -2147483648

Answer: Option A

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.

13.

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";
}
}

A.	slip stream
B.	slipstream stream
C.	stream slip stream
D.	slipstream slip stream

Answer: Option D

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".

14.

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;
}
}

A.	true true
B.	false true
C.	true false
D.	false false

Answer: Option B

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.

15.

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;
}
}

A.	12 15
B.	15 15
C.	3 4 5 3 7 5
D.	3 7 5 3 7 5

Answer: Option B

Explanation:

Output: 15 15

The 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

Operators and Assignments - Finding the output (2)

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