Number literal formats (underscore, hex, binary) in Kotlin

Introduction

Number literals let you write numbers in your code clearly and in different styles. Using underscores, hex, or binary makes numbers easier to read or match real-world formats.

When you want to write large numbers clearly, like 1_000_000 instead of 1000000.

When you need to work with colors or memory addresses, which are often shown in hexadecimal (hex).

When dealing with bits or flags, binary numbers help you see the exact bits set.

When you want to improve code readability by grouping digits logically.

When you want to match number formats used in hardware or protocols.

Syntax

Kotlin

val decimal = 1_000_000
val hex = 0xFF_EC_DE_5E
val binary = 0b1101_0101

Underscores (_) can be placed anywhere between digits to improve readability.

Hex numbers start with 0x and binary numbers start with 0b.

Examples

Underscores make the large number easier to read.

Kotlin

val million = 1_000_000

Hexadecimal number with underscores for clarity, often used for colors.

Kotlin

val color = 0xFF_EC_DE_5E

Binary number with underscores to separate bits visually.

Kotlin

val flags = 0b1101_0101

Sample Program

This program shows how to write and print numbers using underscores, hex, and binary formats.

Kotlin

fun main() {
    val decimal = 1_000_000
    val hex = 0xFF_EC_DE_5E
    val binary = 0b1101_0101

    println("Decimal: $decimal")
    println("Hex: $hex")
    println("Binary: $binary")
}

Important Notes

Underscores cannot be at the start or end of a number, or next to the prefix (0x, 0b).

Hex and binary literals are converted to decimal values when used in calculations or printed.

Summary

Use underscores to make numbers easier to read.

Hex numbers start with 0x and are useful for colors and memory values.

Binary numbers start with 0b and help when working with bits.

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