[ios] How to iterate for loop in reverse order in swift?

When I use the for loop in Playground, everything worked fine, until I changed the first parameter of for loop to be the highest value. (iterated in descending order)

Is this a bug? Did any one else have it?

for index in 510..509
{
    var a = 10
}

The counter that displays the number of iterations that will be executions keeps ticking...

enter image description here

This question is related to ios swift swift-playground

The answer is


as for Swift 2.2 , Xcode 7.3 (10,June,2016) :

for (index,number) in (0...10).enumerate() {
    print("index \(index) , number \(number)")
}

for (index,number) in (0...10).reverse().enumerate() {
    print("index \(index) , number \(number)")
}

Output :

index 0 , number 0
index 1 , number 1
index 2 , number 2
index 3 , number 3
index 4 , number 4
index 5 , number 5
index 6 , number 6
index 7 , number 7
index 8 , number 8
index 9 , number 9
index 10 , number 10


index 0 , number 10
index 1 , number 9
index 2 , number 8
index 3 , number 7
index 4 , number 6
index 5 , number 5
index 6 , number 4
index 7 , number 3
index 8 , number 2
index 9 , number 1
index 10 , number 0

Swift 4.0

for i in stride(from: 5, to: 0, by: -1) {
    print(i) // 5,4,3,2,1
}

If you want to include the to value:

for i in stride(from: 5, through: 0, by: -1) {
    print(i) // 5,4,3,2,1,0
}

For Swift 2.0 and above you should apply reverse on a range collection

for i in (0 ..< 10).reverse() {
  // process
}

It has been renamed to .reversed() in Swift 3.0


With Swift 5, according to your needs, you may choose one of the four following Playground code examples in order to solve your problem.


#1. Using ClosedRange reversed() method

ClosedRange has a method called reversed(). reversed() method has the following declaration:

func reversed() -> ReversedCollection<ClosedRange<Bound>>

Returns a view presenting the elements of the collection in reverse order.

Usage:

let reversedCollection = (0 ... 5).reversed()

for index in reversedCollection {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

As an alternative, you can use Range reversed() method:

let reversedCollection = (0 ..< 6).reversed()

for index in reversedCollection {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

#2. Using sequence(first:next:) function

Swift Standard Library provides a function called sequence(first:next:). sequence(first:next:) has the following declaration:

func sequence<T>(first: T, next: @escaping (T) -> T?) -> UnfoldFirstSequence<T>

Returns a sequence formed from first and repeated lazy applications of next.

Usage:

let unfoldSequence = sequence(first: 5, next: {
    $0 > 0 ? $0 - 1 : nil
})

for index in unfoldSequence {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

#3. Using stride(from:through:by:) function

Swift Standard Library provides a function called stride(from:through:by:). stride(from:through:by:) has the following declaration:

func stride<T>(from start: T, through end: T, by stride: T.Stride) -> StrideThrough<T> where T : Strideable

Returns a sequence from a starting value toward, and possibly including, an end value, stepping by the specified amount.

Usage:

let sequence = stride(from: 5, through: 0, by: -1)

for index in sequence {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

As an alternative, you can use stride(from:to:by:):

let sequence = stride(from: 5, to: -1, by: -1)

for index in sequence {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

#4. Using AnyIterator init(_:) initializer

AnyIterator has an initializer called init(_:). init(_:) has the following declaration:

init(_ body: @escaping () -> AnyIterator<Element>.Element?)

Creates an iterator that wraps the given closure in its next() method.

Usage:

var index = 5

guard index >= 0 else { fatalError("index must be positive or equal to zero") }

let iterator = AnyIterator({ () -> Int? in
    defer { index = index - 1 }
    return index >= 0 ? index : nil
})

for index in iterator {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

If needed, you can refactor the previous code by creating an extension method for Int and wrapping your iterator in it:

extension Int {

    func iterateDownTo(_ endIndex: Int) -> AnyIterator<Int> {
        var index = self
        guard index >= endIndex else { fatalError("self must be greater than or equal to endIndex") }

        let iterator = AnyIterator { () -> Int? in
            defer { index = index - 1 }
            return index >= endIndex ? index : nil
        }
        return iterator
    }

}

let iterator = 5.iterateDownTo(0)

for index in iterator {
    print(index)
}

/*
Prints:
5
4
3
2
1
0
*/

You can consider using the C-Style while loop instead. This works just fine in Swift 3:

var i = 5 
while i > 0 { 
    print(i)
    i -= 1
}

Apply the reverse function to the range to iterate backwards:

For Swift 1.2 and earlier:

// Print 10 through 1
for i in reverse(1...10) {
    println(i)
}

It also works with half-open ranges:

// Print 9 through 1
for i in reverse(1..<10) {
    println(i)
}

Note: reverse(1...10) creates an array of type [Int], so while this might be fine for small ranges, it would be wise to use lazy as shown below or consider the accepted stride answer if your range is large.


To avoid creating a large array, use lazy along with reverse(). The following test runs efficiently in a Playground showing it is not creating an array with one trillion Ints!

Test:

var count = 0
for i in lazy(1...1_000_000_000_000).reverse() {
    if ++count > 5 {
        break
    }
    println(i)
}

For Swift 2.0 in Xcode 7:

for i in (1...10).reverse() {
    print(i)
}

Note that in Swift 2.0, (1...1_000_000_000_000).reverse() is of type ReverseRandomAccessCollection<(Range<Int>)>, so this works fine:

var count = 0
for i in (1...1_000_000_000_000).reverse() {
    count += 1
    if count > 5 {
        break
    }
    print(i)
}

For Swift 3.0 reverse() has been renamed to reversed():

for i in (1...10).reversed() {
    print(i) // prints 10 through 1
}

You can use reversed() method for easily reverse values.

var i:Int
for i in 1..10.reversed() {
    print(i)
}

The reversed() method reverse the values.


For me, this is the best way.

var arrayOfNums = [1,4,5,68,9,10]

for i in 0..<arrayOfNums.count {
    print(arrayOfNums[arrayOfNums.count - i - 1])
}

Swift 4 onwards

for i in stride(from: 5, to: 0, by: -1) {
    print(i)
}
//prints 5, 4, 3, 2, 1

for i in stride(from: 5, through: 0, by: -1) {
    print(i)
}
//prints 5, 4, 3, 2, 1, 0

In Swift 4 and latter

    let count = 50//For example
    for i in (1...count).reversed() {
        print(i)
    }

var sum1 = 0
for i in 0...100{
    sum1 += i
}
print (sum1)

for i in (10...100).reverse(){
    sum1 /= i
}
print(sum1)

Updated for Swift 3

The answer below is a summary of the available options. Choose the one that best fits your needs.

reversed: numbers in a range

Forward

for index in 0..<5 {
    print(index)
}

// 0
// 1
// 2
// 3
// 4

Backward

for index in (0..<5).reversed() {
    print(index)
}

// 4
// 3
// 2
// 1
// 0

reversed: elements in SequenceType

let animals = ["horse", "cow", "camel", "sheep", "goat"]

Forward

for animal in animals {
    print(animal)
}

// horse
// cow
// camel
// sheep
// goat

Backward

for animal in animals.reversed() {
    print(animal)
}

// goat
// sheep
// camel
// cow
// horse

reversed: elements with an index

Sometimes an index is needed when iterating through a collection. For that you can use enumerate(), which returns a tuple. The first element of the tuple is the index and the second element is the object.

let animals = ["horse", "cow", "camel", "sheep", "goat"]

Forward

for (index, animal) in animals.enumerated() {
    print("\(index), \(animal)")
}

// 0, horse
// 1, cow
// 2, camel
// 3, sheep
// 4, goat

Backward

for (index, animal) in animals.enumerated().reversed()  {
    print("\(index), \(animal)")
}

// 4, goat
// 3, sheep
// 2, camel
// 1, cow
// 0, horse

Note that as Ben Lachman noted in his answer, you probably want to do .enumerated().reversed() rather than .reversed().enumerated() (which would make the index numbers increase).

stride: numbers

Stride is way to iterate without using a range. There are two forms. The comments at the end of the code show what the range version would be (assuming the increment size is 1).

startIndex.stride(to: endIndex, by: incrementSize)      // startIndex..<endIndex
startIndex.stride(through: endIndex, by: incrementSize) // startIndex...endIndex

Forward

for index in stride(from: 0, to: 5, by: 1) {
    print(index)
}

// 0
// 1
// 2
// 3
// 4

Backward

Changing the increment size to -1 allows you to go backward.

for index in stride(from: 4, through: 0, by: -1) {
    print(index)
}

// 4
// 3
// 2
// 1
// 0

Note the to and through difference.

stride: elements of SequenceType

Forward by increments of 2

let animals = ["horse", "cow", "camel", "sheep", "goat"]

I'm using 2 in this example just to show another possibility.

for index in stride(from: 0, to: 5, by: 2) {
    print("\(index), \(animals[index])")
}

// 0, horse
// 2, camel
// 4, goat

Backward

for index in stride(from: 4, through: 0, by: -1) {
    print("\(index), \(animals[index])")
}

// 4, goat
// 3, sheep 
// 2, camel
// 1, cow  
// 0, horse 

Notes


If one is wanting to iterate through an array (Array or more generally any SequenceType) in reverse. You have a few additional options.

First you can reverse() the array and loop through it as normal. However I prefer to use enumerate() much of the time since it outputs a tuple containing the object and it's index.

The one thing to note here is that it is important to call these in the right order:

for (index, element) in array.enumerate().reverse()

yields indexes in descending order (which is what I generally expect). whereas:

for (index, element) in array.reverse().enumerate() (which is a closer match to NSArray's reverseEnumerator)

walks the array backward but outputs ascending indexes.


Examples related to ios

Adding a UISegmentedControl to UITableView Crop image to specified size and picture location Undefined Symbols error when integrating Apptentive iOS SDK via Cocoapods Keep placeholder text in UITextField on input in IOS Accessing AppDelegate from framework? Autoresize View When SubViews are Added Warp \ bend effect on a UIView? Speech input for visually impaired users without the need to tap the screen make UITableViewCell selectable only while editing Xcode 12, building for iOS Simulator, but linking in object file built for iOS, for architecture arm64

Examples related to swift

Make a VStack fill the width of the screen in SwiftUI Xcode 10.2.1 Command PhaseScriptExecution failed with a nonzero exit code Command CompileSwift failed with a nonzero exit code in Xcode 10 Convert Json string to Json object in Swift 4 iOS Swift - Get the Current Local Time and Date Timestamp Xcode 9 Swift Language Version (SWIFT_VERSION) How do I use Safe Area Layout programmatically? How can I use String substring in Swift 4? 'substring(to:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator Safe Area of Xcode 9 The use of Swift 3 @objc inference in Swift 4 mode is deprecated?

Examples related to swift-playground

Class 'ViewController' has no initializers in swift How to iterate for loop in reverse order in swift? How to print to console using swift playground?