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LESSON 56 OF 348 · STRINGS I

Accessing Characters in a Kotlin String

Lesson Read the full explanation

Each character in a string has a numbered position known as its index. We access a character by referring to its index number.

Strings use zero-based indexing, so the first character in a string has an index of 0, the second character has an index of 1, etc.

We’re using the .charAt() method with the index to access the character:

val message = "Hello world!"

var c = message[0]
println(c) // Output: H

c = message[1]
println(c) // Output: e

println(message[6]) // Output: w

Whitespaces have indices:

Note that every character inside the quotes (including whitespaces) is part of the string and has its own index. For example, in string "Hello world!", the whitespace is placed on index 5.

You can see that clearly when you run this code:

val message = "Hello world!"

println("${message[4]}${message[5]}${message[6]}")

the output is:

o w

Assignment
Follow the Coding Tutorial and let's access some characters!

Hint
Look at the examples above if you get stuck.

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    KOTLIN · STRINGS I

    Accessing Characters in Kotlin, explained

    Indexes start at zero

    Every character in a string has a numbered position, its index. Kotlin counts from zero, so the first character is at index 0, the second at index 1, and so on. You read a character by writing the string's name followed by the index in square brackets:

    val message = "Hello world!"

var c = message[0]
println(c) // H

c = message[1]
println(c) // e

println(message[6]) // w

    The brackets are a shorthand for the string's get function: message[6] is the same as message.get(6). Most of the time you'll use the brackets.

    Whitespace counts

    Every character between the quotes is part of the string, including spaces. In "Hello world!", the space sits at index 5, between o at index 4 and w at index 6:

    val message = "Hello world!"

println("${message[4]}${message[5]}${message[6]}")

    o w

    The ${...} parts are string templates, which put the value of an expression into the text. Here each one inserts a single character.

    What you get back is a Char

    Indexing a string doesn't give you a one-character string. It gives you a Char, a separate type for a single character. A Char literal is written with single quotes, as in 'a', while a string uses double quotes, as in "a". The difference matters as soon as you do arithmetic on the result:

    val fruit = "Avocado"
val c = fruit[1] // 'v'

println(c + 1)         // w
println(c.code)        // 118
println(c.toString() + "x") // vx

    Adding an integer to a Char gives the character that many places further along in the character table, so 'v' + 1 is 'w'. The property code is the character's numeric value. To join a Char to text, turn it into a string first with toString(), or use a template.

    Printing a Char with println shows the character itself, so no conversion is needed to print it.

    Counting from the end

    The last index is always one less than the length, because counting starts at 0. In "Avocado" there are 7 characters, so the indexes run from 0 to 6, and the last character is fruit[6]. Using the length keeps the code correct for any string:

    val fruit = "Avocado"

println(fruit[fruit.length - 1]) // o
println(fruit.last())            // o
println(fruit.first())           // A
println(fruit.lastIndex)         // 6

    Kotlin has no negative indexes. There is no fruit[-1] to ask for the last character.

    Visiting every character

    Later you'll often want every index in turn. The property indices gives exactly the valid ones, from 0 up to the last, so a loop never goes out of range:

    val fruit = "Avocado"

for (i in fruit.indices) {
    println("$i: ${fruit[i]}")
}

    This prints seven lines, from 0: A to 6: o. When the position doesn't matter, a loop can take the characters directly: for (ch in fruit) gives each Char in order, and comparing one to a character literal such as 'o' finds the two letters o in this word.

    Gotchas

    Going past the end

    An index that doesn't exist makes the program stop with an exception. For "Avocado", both fruit[7] and fruit[-1] throw a StringIndexOutOfBoundsException, and the message reads Index 7 out of bounds for length 7 (or Index -1 ...). If the index may be invalid, fruit.getOrNull(10) returns null instead of throwing, and fruit.getOrElse(10) { '?' } returns the fallback you give it.

    Strings can't be changed in place

    You can read a character by index, but you can't assign to one. The line fruit[0] = 'a' doesn't compile: the compiler reports no 'set' operator method providing array access. To get a modified text, build a new string.

    The exercise, step by step

    The starting code defines val fruit = "Avocado".

    1. Print the first character. Use index 0: println(fruit[0]). It prints A.
    2. Print the last character. The string has 7 characters, so the last index is 6: println(fruit[6]). It prints o.
    val fruit = "Avocado"

println(fruit[0])
println(fruit[6])

    The output:

    A
o

    The two lines fruit[fruit.length - 1] and fruit.last() would give the same o.

    Common mistakes

    • Starting at 1. fruit[1] is the second character, v.
    • Using the length as an index. fruit[fruit.length] is one past the end and throws.
    • Using parentheses. fruit(0) is not valid; indexing uses square brackets.
    • Mixing up 'a' and "a". Single quotes make a Char and double quotes make a String, and the two are not interchangeable.

    Key takeaways

    • Indexes start at 0, and spaces have indexes too.
    • text[i] gives a Char.
    • The last index is length - 1, also available as lastIndex; last() returns the last character directly.
    • An index out of range throws an exception, and strings can't be modified by index.
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