Unary operators in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to increment the variable count by 1 using a unary operator.

Java

int count = 5;
count[1];

Drag options to blanks, or click blank then click option'

A-

B++

C+=

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '--' which decreases the value.

Using '+=' which is not a unary operator.

Using '-' which subtracts but is not unary increment.

The unary increment operator ++ increases the value of count by 1.

2fill in blank

medium

Complete the code to negate the boolean variable isActive using a unary operator.

Java

boolean isActive = true;
isActive = [1]isActive;

Drag options to blanks, or click blank then click option'

A!

B--

C++

D~

Attempts:

3 left

💡 Hint

Common Mistakes

Using '++' or '--' which are for numeric increments or decrements.

Using '~' which is bitwise complement, not for booleans.

The unary logical complement operator ! negates a boolean value.

3fill in blank

hard

Fix the error in the code by choosing the correct unary operator to decrement num by 1.

Java

int num = 10;
num[1];

Drag options to blanks, or click blank then click option'

A*

B++

C+

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '++' which increases the value.

Using '+' or '*' which are binary operators and cause errors here.

The unary decrement operator -- decreases the value of num by 1.

4fill in blank

hard

Fill both blanks to create a map of numbers to their negations using unary operators.

Java

Map<Integer, Integer> negations = Map.of(1, [1]1, 2, [2]2);

Drag options to blanks, or click blank then click option'

A-

B++

C--

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using '++' or '--' which change the number by one, not negate.

Using '!' which is for booleans, not numbers.

The unary minus operator - negates numbers, turning positive to negative.

5fill in blank

hard

Fill all three blanks to create a map where keys are numbers and values are their incremented, negated, and decremented forms respectively.

Java

int one = 1, two = 2, three = 3;
Map<Integer, Integer> results = Map.of(
  one, [1]one,
  two, [2]two,
  three, [3]three);

Drag options to blanks, or click blank then click option'

A++

B-

C--

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!' which is only for booleans.

Mixing prefix and postfix operators incorrectly.

Use ++ to increment, - to negate, and -- to decrement the numbers.

Practice

1.

What does the unary operator ++ do in Java?

easy

A. Negates the value of a variable

B. Increases the value of a variable by 1

C. Decreases the value of a variable by 1

D. Checks if a value is true or false

Solution

1. Step 1: Understand the meaning of ++

   The ++ operator adds 1 to the current value of a variable.

2. Step 2: Compare with other unary operators

   -- subtracts 1, ! flips boolean, and negation changes sign.

3. Final Answer:

   Increases the value of a variable by 1 -> Option B

4. Quick Check:

   ++ means add one [OK]

Hint: ++ means add one to a number quickly [OK]

Common Mistakes:

• Confusing ++ with --
• Thinking ++ negates the value
• Mixing ++ with boolean negation !

2.

Which of the following is the correct **postfix** syntax to decrement a variable x by 1 using a unary operator?

int x = 5;
// Your code here

easy

A. x-1;

B. --x;

C. x--;

D. x=!1;

Solution

1. Step 1: Identify decrement unary operators

   Both x-- and --x decrement by 1, but x-1 is an expression, not an operator.

2. Step 2: Choose correct syntax for decrement

   x--; correctly decreases x by 1 after its current use.

3. Final Answer:

   x--; -> Option C

4. Quick Check:

   Use x-- to subtract one [OK]

Hint: Use x-- or --x to subtract one [OK]

Common Mistakes:

• Using x-1; which doesn't change x
• Confusing ! operator with decrement
• Writing invalid syntax like x=!1;

3.

What is the output of the following Java code?

int a = 3;
System.out.println(++a);
System.out.println(a++);
System.out.println(a);

medium

A. 4 4 5

B. 3 4 5

C. 4 3 4

D. 3 3 4

Solution

1. Step 1: Evaluate ++a

   ++a increases a before printing, so prints 4 and a becomes 4.

2. Step 2: Evaluate a++

   a++ prints current a (4) then increases it to 5.

3. Step 3: Print final value of a

   After previous steps, a is 5, so prints 5.

4. Final Answer:

   4 4 5 -> Option A

5. Quick Check:

   Prefix ++ adds before print, postfix ++ adds after print [OK]

Hint: Prefix ++ changes before print, postfix ++ after print [OK]

Common Mistakes:

• Mixing prefix and postfix effects
• Forgetting that ++ changes the variable
• Assuming no change in variable after print

4.

Find the error in the following code snippet:

boolean flag = true;
flag = !flag++;
System.out.println(flag);

medium

A. No error, code runs fine

B. Missing semicolon after flag declaration

C. Logical NOT operator ! cannot be used here

D. Cannot apply ++ to a boolean variable

Solution

1. Step 1: Check variable type and operators

   flag is boolean; ++ operator works only on numeric types, not boolean.

2. Step 2: Identify invalid operation

   Using flag++ causes a compile-time error because booleans cannot be incremented.

3. Final Answer:

   Cannot apply ++ to a boolean variable -> Option D

4. Quick Check:

   ++ works only on numbers, not booleans [OK]

Hint: ++ cannot be used on boolean variables [OK]

Common Mistakes:

• Trying to increment boolean values
• Confusing logical NOT ! with increment ++
• Assuming code compiles without error

5.

Given the code below, what will be the final value of count after execution?

int count = 0;
for (int i = 0; i < 3; i++) {
    count = count++ + ++count;
}
System.out.println(count);

hard

A. 14

B. 6

C. 12

D. 9

Solution

1. Step 1: Understand count = count++ + ++count; behavior

   Binary + evaluates left operand first (count++: old value then increment), then right (++count: increment then new value).

2. Step 2: Trace loop iterations

   Iteration 1 (count=0): count++ value=0 (count=1), ++count value=2, sum=2 -> count=2.
   Iteration 2 (count=2): count++=2 (count=3), ++count=4, sum=6 -> count=6.
   Iteration 3 (count=6): count++=6 (count=7), ++count=8, sum=14 -> count=14.

3. Step 3: Final value after loop

   After 3 iterations, count is 14.

4. Final Answer:

   14 -> Option A

5. Quick Check:

   Left-to-right: postfix old + prefix new [OK]

Hint: Left operand postfix (old then inc), right prefix (inc then new) [OK]

Common Mistakes:

• Assuming both increments add fully
• Ignoring postfix increment side effects
• Calculating sum without assignment effect