Number literal formats (underscore, hex, binary) in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an integer with underscores for readability.

Kotlin

val number = 1[1]000_000

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

A_

B-

C.

D,

Attempts:

3 left

💡 Hint

Common Mistakes

Using commas or dots inside number literals causes syntax errors.

Trying to use hyphens '-' inside numbers is invalid.

In Kotlin, underscores can be used inside number literals to improve readability, like 1_000_000.

2fill in blank

medium

Complete the code to declare a hexadecimal number literal.

Kotlin

val hexNumber = [1]FF

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

A0x

B0o

C0d

D0b

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0b which is for binary numbers.

Using 0d or 0o which are invalid prefixes in Kotlin.

Hexadecimal literals in Kotlin start with 0x followed by hex digits.

3fill in blank

hard

Fix the error in the binary number literal declaration.

Kotlin

val binaryNumber = [1]1010_1010

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

A0d

B0x

C0b

D0o

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0x which is for hexadecimal numbers.

Using 0d or 0o which are invalid prefixes.

Binary literals in Kotlin start with 0b followed by binary digits.

4fill in blank

hard

Fill both blanks to create a hexadecimal number with underscores.

Kotlin

val color = [1]FF[2]00FF

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

A0x

B_

C0b

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using hyphens instead of underscores.

Using binary prefix 0b for hex numbers.

Hexadecimal literals start with 0x. Underscores can separate digits for readability.

5fill in blank

hard

Fill all three blanks to declare a binary number with underscores and the correct prefix.

Kotlin

val mask = [1]1010[2]1010[3]1111

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

A0b

B_

C0x

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using hexadecimal prefix 0x for binary numbers.

Using hyphens instead of underscores.

Binary literals start with 0b. Underscores _ can separate digits for clarity.

Practice

1. Which of the following is a valid way to write the number one million in Kotlin using underscores for readability?

easy

A. 1__000000

B. 1000_000_

C. 10__00_000

D. 1_000_000

Solution

1. Step 1: Understand underscore usage in Kotlin numbers

   Kotlin allows underscores between digits to improve readability, but they must not be consecutive, leading, or trailing.

2. Step 2: Check each option for correct underscore placement

   1_000_000 uses single underscores separating groups of three digits correctly. A has consecutive underscores (invalid). B has trailing underscore (invalid). C has consecutive underscores (invalid).

3. Final Answer:

   1_000_000 -> Option D

4. Quick Check:

   Underscores separate digits properly = 1_000_000 [OK]

Hint: Use single underscores between digit groups only [OK]

Common Mistakes:

• Using double underscores
• Placing underscores at start or end
• Trailing underscores

2. Which of the following Kotlin number literals is NOT valid syntax?

easy

A. 123_456_

B. 0b1010_1101

C. 0x1A3F

D. 987_654

Solution

1. Step 1: Review Kotlin rules for underscores in number literals

   Underscores can be used between digits but not at the end or start of a number.

2. Step 2: Check each option for trailing underscores

   123_456_ ends with an underscore, which is invalid syntax. Options A, B, and D are valid hexadecimal, binary, and decimal literals respectively.

3. Final Answer:

   123_456_ -> Option A

4. Quick Check:

   Trailing underscore invalid = 123_456_ [OK]

Hint: Never put underscore at the end of a number [OK]

Common Mistakes:

• Trailing underscores in numbers
• Confusing hex prefix 0x with decimal
• Misplacing underscores in binary literals

3. What is the output of this Kotlin code?

val x = 0xF_A
val y = 0b1010_1010
println(x + y)

medium

A. 270

B. 250

C. 420

D. 280

Solution

1. Step 1: Convert hexadecimal and binary literals to decimal

   0xF_A means hex FA: F=15, A=10, so 15*16 + 10 = 250. 0b1010_1010 is binary 10101010 which equals 128+32+8+2=170.

2. Step 2: Add the decimal values

   250 + 170 = 420.

3. Final Answer:

   420 -> Option C

4. Quick Check:

   (0xFA + 0b10101010) = 420 [OK]

Hint: Convert hex and binary to decimal before adding [OK]

Common Mistakes:

• Misreading hex digits
• Incorrect binary to decimal conversion
• Ignoring underscores in literals

4. Identify the error in this Kotlin code snippet:

val num = 0b1021_0010

medium

A. Underscores are not allowed in binary literals

B. Binary literals cannot contain digits other than 0 or 1

C. Binary literals must start with 0x

D. The number is too large for binary format

Solution

1. Step 1: Understand valid digits in binary literals

   Binary literals can only contain digits 0 and 1, optionally separated by underscores.

2. Step 2: Check the given literal for invalid digits

   The literal contains '2', which is invalid in binary. Underscores are allowed, and binary literals start with 0b, not 0x.

3. Final Answer:

   Binary literals cannot contain digits other than 0 or 1 -> Option B

4. Quick Check:

   Binary digits must be 0 or 1 only [OK]

Hint: Binary digits only 0 or 1, no other digits allowed [OK]

Common Mistakes:

• Using digits other than 0 or 1 in binary
• Confusing hex prefix 0x with binary 0b
• Thinking underscores are disallowed in binary

5. You want to represent the decimal number 255 in Kotlin using a binary literal with underscores for readability. Which of the following is the best way to write it?

hard

A. 0b1111_1111

B. 0b11111111

C. 0b1111_1110

D. 0b1111_11112

Solution

1. Step 1: Convert decimal 255 to binary

   255 in binary is 11111111 (eight ones).

2. Step 2: Add underscores for readability

   Grouping bits in fours is common, so 1111_1111 is clearer and valid.

3. Step 3: Check options for correctness

   0b1111_1111 correctly uses underscores and matches 255. 0b11111111 is correct but less readable. 0b1111_1110 is 254, and 0b1111_11112 has invalid digit '2'.

4. Final Answer:

   0b1111_1111 -> Option A

5. Quick Check:

   Binary 1111_1111 = 255 decimal [OK]

Hint: Group binary digits in fours with underscores [OK]

Common Mistakes:

• Using wrong binary digits
• Adding invalid digits like 2
• Not grouping bits for readability