Reverse Range in Swift

Question

Is there a way to work with reverse ranges in Swift?

For example:

for i in 5...1 {
  // do something
}

is an infinite loop.

Answer 1

It appears that the answers to this question have changed a bit as we've progressed through the betas. As of beta 4, both the by() function and the ReversedRange type have been removed from the language. If you're looking to make a reversed range, your options are now as follows:

1: Create a forward range, and then use the reverse() function to reverse it.

for x in reverse(0 ... 4) {
    println(x) // 4, 3, 2, 1, 0
}

for x in reverse(0 ..< 4) {
    println(x) // 3, 2, 1, 0
}

2: Use the new stride() functions that were added in beta 4, which includes functions to specify the starting and ending indexes, as well as the amount to iterate by.

for x in stride(from: 0, through: -8, by: -2) {
    println(x) // 0, -2, -4, -6, -8
}

for x in stride(from: 6, to: -2, by: -4) {
    println(x) // 6, 2
}

Note that I've also included the new exclusive range operator in this post as well. .. was replaced with ..<.

Edit: From the Xcode 6 beta 5 release notes, Apple added the following suggestion for handling this:

ReverseRange has been removed; use lazy(x..

Here's an example.

for i in lazy(0...5).reverse() {
    // 0, 1, 2, 3, 4, 5
}

Answer 2

Reverse() function is used for reverse number.

Var n:Int // Enter number

For i in 1...n.reverse() { Print(i) }

Answer 3

Swift 3, 4+: you can do it like this:

for i in sequence(first: 10, next: {$0 - 1}) {

    guard i >= 0 else {
        break
    }
    print(i)
}

result: 10, 9, 8 ... 0

You can customise it any way you like. For more info read func sequence<T> reference

Answer 4

This could be another way of doing this.

(1...5).reversed().forEach { print($0) }

Answer 5

There's something troubling about the asymmetry of this:

for i in (1..<5).reverse()

...as opposed to this:

for i in 1..<5 {

It means that every time I want to do a reverse range, I have to remember to put the parentheses, plus I have to write that .reverse() on the end, sticking out like a sore thumb. This is really ugly in comparison to C-style for loops, which are symmetrical counting up and counting down. So I tended to use C-style for loops instead. But in Swift 2.2, C-style for loops are going away! So I've had to scurry around replacing all my decrementing C-style for loops with this ugly .reverse() construct — wondering all the while, why on earth isn't there a reverse-range operator?

But wait! This is Swift — we're allowed to define our own operators!! Here we go:

infix operator >>> {
    associativity none
    precedence 135
}

func >>> <Pos : ForwardIndexType where Pos : Comparable>(end:Pos, start:Pos)
    -> ReverseRandomAccessCollection<(Range<Pos>)> {
        return (start..<end).reverse()
}

So now I'm allowed to say:

for i in 5>>>1 {print(i)} // 4, 3, 2, 1

This covers just the most common case that occurs in my code, but it is far and away the most common case, so it's all I need at present.

I had a kind of internal crisis coming up with the operator. I would have liked to use >.., as being the reverse of ..<, but that's not legal: you can't use a dot after a non-dot, it appears. I considered ..> but decided it was too hard to distinguish from ..<. The nice thing about >>> is that it screams at you: "down to!" (Of course you're free to come up with another operator. But my advice is: for super symmetry, define <<< to do what ..< does, and now you've got <<< and >>> which are symmetrical and easy to type.)

---

Swift 3 version (Xcode 8 seed 6):

infix operator >>> : RangeFormationPrecedence
func >>><Bound>(maximum: Bound, minimum: Bound) ->
    ReversedRandomAccessCollection<CountableRange<Bound>> 
    where Bound : Comparable, Bound.Stride : Integer {
        return (minimum..<maximum).reversed()
}

Swift 4 version (Xcode 9 beta 3):

infix operator >>> : RangeFormationPrecedence
func >>><Bound>(maximum: Bound, minimum: Bound)
    -> ReversedRandomAccessCollection<CountableRange<Bound>>
    where Bound : Comparable & Strideable { 
        return (minimum..<maximum).reversed()
}

Swift 4.2 version (Xcode 10 beta 1):

infix operator >>> : RangeFormationPrecedence
func >>><Bound>(maximum: Bound, minimum: Bound)
    -> ReversedRandomAccessCollection<Range<Bound>>
    where Bound : Strideable { 
        return (minimum..<maximum).reversed()
}

Swift 5 version (Xcode 10.2.1):

infix operator >>> : RangeFormationPrecedence
func >>><Bound>(maximum: Bound, minimum: Bound)
    -> ReversedCollection<Range<Bound>>
    where Bound : Strideable {
        return (minimum..<maximum).reversed()
}

Answer 6

Update For latest Swift 3 (still works in Swift 4)

You can use the reversed() method on a range

for i in (1...5).reversed() { print(i) } // 5 4 3 2 1

Or stride(from:through:by:) method

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

stide(from:to:by:) is similar but excludes the last value

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

Update For latest Swift 2

First of all, protocol extensions change how reverse is used:

for i in (1...5).reverse() { print(i) } // 5 4 3 2 1

Stride has been reworked in Xcode 7 Beta 6. The new usage is:

for i in 0.stride(to: -8, by: -2) { print(i) } // 0 -2 -4 -6
for i in 0.stride(through: -8, by: -2) { print(i) } // 0 -2 -4 -6 -8

It also works for Doubles:

for i in 0.5.stride(to:-0.1, by: -0.1) { print(i) }

Be wary of floating point compares here for the bounds.

Earlier edit for Swift 1.2: As of Xcode 6 Beta 4, by and ReverseRange don't exist anymore :[

If you are just looking to reverse a range, the reverse function is all you need:

for i in reverse(1...5) { println(i) } // prints 5,4,3,2,1

As posted by 0x7fffffff there is a new stride construct which can be used to iterate and increment by arbitrary integers. Apple also stated that floating point support is coming.

Sourced from his answer:

for x in stride(from: 0, through: -8, by: -2) {
    println(x) // 0, -2, -4, -6, -8
}

for x in stride(from: 6, to: -2, by: -4) {
    println(x) // 6, 2
}