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LESSON 98 OF 415 · ARRAYS I

Updating Array Elements in Swift

Lesson Read the full explanation

The entries of Arrays are mutable and can be changed freely, using indices and subscript notation:

Assigning to array[index] replaces the element placed at that index with the new value:

var numbers = [50, 40, 30]

numbers[0] = 15
numbers[2] = -1

// numbers is now [15, 40, -1]

Assignment
Follow the Coding Tutorial and let's play with some arrays.

Hint
Look at the examples above if you get stuck.

STEP 1 OF 2
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    SWIFT · ARRAYS I

    Updating Elements in Swift, explained

    Subscript assignment

    The entries of a Swift array are mutable: you can replace any of them using its index. Write the array's name, the index in square brackets, then = and the new value.

    var numbers = [50, 40, 30]

numbers[0] = 15
numbers[2] = -1

print(numbers)  // [15, 40, -1]

    Two things are involved: the index says which slot to change, and the value says what to store there. The slot at index 1 is untouched, and the length of the array stays 3.

    Requirements

    • The array is declared with var. A let array can't be changed, so fruits[0] = "apple" on a constant fails to compile with a message that the array is a let constant.
    • The index exists. The subscript replaces an existing element. Assigning to an index equal to or beyond count crashes the program, and it does not grow the array. To add elements, use append.
    • The value has the element type. You can store an Int in an [Int], but not a String or a Double.

    Updating with the current value

    To change an element relative to its old value, use a compound operator. It reads the current value, applies the operation and stores the result in the same slot:

    var ages = [14, 32, 50, 21]
ages[1] += 5
print(ages)  // [14, 37, 50, 21]

    The line is shorthand for ages[1] = ages[1] + 5. -=, *= and the other compound operators work the same way. Strings have += too, which appends text: fruits[0] += "s" would turn "mango" into "mangos".

    Changing every element

    A loop over the valid indexes can update each element in place:

    var scores = [14, 37, 50, 21]
for i in 0..<scores.count {
    scores[i] += 1
}
print(scores)  // [15, 38, 51, 22]

    Looping with for score in scores and writing score += 1 does not compile. The loop variable is a constant copy of each element, so it can't be used to change the array.

    Swapping two elements

    Subscript assignment is also how you exchange two values. Keep one of them in a temporary constant first, or the first assignment would destroy it:

    var a = [3, 1, 2]
let tmp = a[0]
a[0] = a[2]
a[2] = tmp
print(a)  // [2, 1, 3]

    Arrays also have swapAt(_:_:), which does the same in one call: a.swapAt(0, 1) turns [2, 1, 3] into [1, 2, 3].

    Changes don't leak into copies

    An array is a value type. After let backup = ages, the two names hold separate arrays, and a later ages[0] = 0 doesn't alter backup.

    The exercise, step by step

    1. Change the first fruit. The first item sits at index 0, so assign "apple" to it: fruits[0] = "apple". The string needs double quotes.
    2. Add 5 to the second age. Index 1 holds 32. Use ages[1] += 5, and the element becomes 37.
    var fruits = ["mango", "kiwi", "grapefruit"]
var ages = [14, 32, 50, 21]

fruits[0] = "apple"
ages[1] += 5

    Printing the arrays afterwards gives ["apple", "kiwi", "grapefruit"] and [14, 37, 50, 21].

    Common mistakes

    • Changing a let array. Switch the declaration to var.
    • Off-by-one indexes. The first item is index 0 and the second is index 1. fruits[1] = "apple" replaces "kiwi".
    • Replacing instead of adding. ages[1] = 5 overwrites 32 with 5, while ages[1] += 5 gives 37.
    • Using a quoted index. The index is a number: fruits[0], not fruits["0"].

    Key takeaways

    • array[index] = value replaces one element and needs a var array.
    • array[index] += n changes an element based on its current value.
    • The index must exist and the value must match the element type.
    • Use append to make an array longer; a subscript can't.
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