Ch3 Operators

Question 1

What is the data type of x * y?

short x = 10;
short y = 3;
var z = x * y;

Answer 1

x and y will both be promoted to int before the binary multiplication operation, resulting in an output type of int.

To have an output type of short again, use casting:
var z = (short) x * y;

Question 2

What is the data type of w * x / y?

short w = 14;
float x = 13;
double y = 30;
var z = w * x / y;

Answer 2

First, w will automatically be promoted to an int because it is a short and it is being used in an arithmetic binary operation.

The promoted w value will then be promoted to a float so it can be multiplied with x.

The result of w * x will then be promoted to a double so it can be divided by y, resulting in a double value.

Question 3

What is the data type of y?

short x = 10;
var y = x++;

Answer 3

Unary operators are excluded from the numeric promotion rules, resulting in a short data type.

Question 4

Does the following code compile?

short mouse = 10;
short hamster = 3;
short capybara = mouse * hamster;

Answer 4

Short values are automatically promoted to in when applying any arithmetic operator, with the resulting value being an int. The code does not compile.

Question 5

What is the value of wolf and coyote? why?

long wolf = 5;
long coyote = (wolf = 3);
System.out.println(wolf);
System.out.println(coyote);

Answer 5

• wolf = 3;
• coyote = 3;
  The result of an assignment operator is an expression in and of itself, equal to the value of the assignment. wolf = 3 also returns the value 3.

Question 6

What does the binary and equality operator ‘==’ do when applied to objects?

Answer 6

Returns true if the two values reference the same object.

Question 7

What does the binary and equality operator ‘!=’ do when applied to objects?

Answer 7

Returns true if the two values do not reference the same object.

Question 8

True or false: The following code returns true:

5 == 5.00

Answer 8

If the numeric values are of different data types, the values are automatically promoted.

Question 9

What is the output of this code? why?

System.out.println(null == null);

Answer 9

In Java, comparing null with null results in true.

Question 10

What is the output of this code? why?

System.out.println(null instanceof Object);

Answer 10

Calling instanceof on the null literal or any null reference always returns false.

Question 11

What is the output of this code? why?

System.out.println(null instanceof null);

Answer 11

This code does not compile because null is used on the right side of the instanceof operator.

Question 12

What change, when applied independently, would allow the following code to compile?

3: long ear = 10;
4: int hearing = 2 * ear;

A) No change.
B) Cast ear on line 4 to int.
C) Change the data type of ear on line 3 to short.
D) Cast 2 * ear on line 4 to int.
E) Change the data type of hearing on line 4 to short.
F) Change the data type of hearing on line 4 to long.

Answer 12

B, C, D, F

Question 13

What is the output of the following code snippet?

4: int pig = (short) 4;
5: pig = pig++;
6: long goat = (int) 2;
7: goat -= 1.0;
8: System.out.println(pig + “ - “ + goat);

A) 4 - 1
B) 4 - 2
C) 5 - 1
D) 5 - 2
E) The code does not compile
F) None of the above

Answer 13

A

Question 14

What is the output of the following code?

public class MyClass {
static long test(double fruit, float vegetable) {
return (int) fruit + vegetable;
}
public static void main(String args[]) {
int y = 4;
int x = (y = 1);
System.out.println(test((long)x, (float) y));
}
}

Answer 14

The code cannot be compiled because the cast ‘(int) fruit + vegetable’ only casts fruit to int but not the whole expression.

Question 15

What is the output of the following code?

short height = 1, weight = 3;
short zebra = (byte) weight * (byte) height;
double ox = 1 + height * 2 + weight;
long giraffe = 1 + 9 % height + 1;
System.out.println(zebra);
System.out.println(ox);
System.out.println(giraffe);

Answer 15

The code does not compile because of the second line. multiplying weight and height will result in an int, which does not fit in a short, unless casted.

Question 16

The following code snippet will print ‘true’.

short s = Short.MAX_VALUE;
char c = s;
System.out.println( c == Short.MAX_VALUE);

Answer 16

This will not compile because a short “variable” can never be assigned to a char without explicit casting. A short “constant” i.e. a short variable defined as final, can be assigned to a char only if the value fits into a char.

short s = 1;
byte b = s; => this will also not compile because although value is small enough to be held by a byte but the Right Hand Side i.e. s is a variable and not a constant.

final short s = 1;
byte b = s; => This is fine because s is a constant and the value fits into a byte.

final short s = 200;
byte b = s; => This is invalid because although s is a constant but the value does not fit into a byte.

Implicit narrowing occurs only for byte, char, short, and int. Remember that it does not occur for long, float, or double. So, this will not compile: int i = 129L; However, an implicit widening conversion from long to a float or a double is valid:

long l = 10L;
float f = l; //valid
double d = l; //valid

Question 17

What will the following code print?

public class TestClass{
static char ch;
static float f;
static boolean bool;
public static void main(String[] args){
System.out.print(f);
System.out.print(“ “);
System.out.print(ch);
System.out.print(“ “);
System.out.print(bool);
}
}

A. 0.0 false
B. 0.0ffalse
C. 0.0 0 false
D. 0.0true
E. 0.0ftrue
F. 0.0f 0 true

Answer 17

A.

This question tests you on two aspects -

the default values that are given to variables of various primitive types. You should remember that all numeric types, including char, get the value of 0 (or 0.0 for float and double) and boolean gets the value of false.
how the value is printed by System.out.print method - java.lang.System class has a public static variable named out, which is of class java.io.PrintStream.The PrintStream class has multiple print/println methods for printing out primitive values as well as objects.
For byte, short, and int, these print/println methods simply print the integer value as it is.

For char, the print/println methods translate the character into one or more bytes according to the platform’s default character encoding. That is why while printing a char value of 0, a blank space is printed instead of 0 (even though the char’s integral value is 0).

For long, float, and double values, these print/println methods use the respective primitive wrapper class’s toString method to get the String representation of that primitive. For example, to print the value of a float variable f, it internally calls Float.toString(f). Now, this method doesn’t append an “f” at the end of a float value. That is why a float value of 0.0 is printed as 0.0 and not 0.0f.