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;
}
}
Answer: B
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
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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.println(i);
}
}
Answer: B
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.
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What will be the output of the program?
class SC2
{
public static void main(String [] args)
{
SC2 s = new SC2();
s.start();
}
void start()
{
int a = 3;
int b = 4;
System.out.print(" " + 7 + 2 + " ");
System.out.print(a + b);
System.out.print(" " + a + b + " ");
System.out.print(foo() + a + b + " ");
System.out.println(a + b + foo());
}
String foo()
{
return "foo";
}
}
Answer: D
Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.
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What will be the output of the program?
class Test
{
public static void main(String [] args)
{
int x=20;
String sup = (x < 15>
Answer: B
This is an example of a nested ternary operator. The second evaluation (x < 22> is true, so the "tiny" value is assigned to sup.
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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> 2 ) || (++y > 2))
{
x++;
}
}
System.out.println(x + " " + y);
}
}
Answer: B
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.
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