问题

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...

$\"enter$

回答1:

Xcode 6 beta 4 added two functions to iterate on ranges with a step other than one: stride(from: to: by:), which is used with exclusive ranges and stride(from: through: by:), which is used with inclusive ranges.

To iterate on a range in reverse order, they can be used as below:

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

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

Note that neither of those is a Range member function. They are global functions that return either a StrideTo or a StrideThrough struct, which are defined differently from the Range struct.

A previous version of this answer used the by() member function of the Range struct, which was removed in beta 4. If you want to see how that worked, check the edit history.

回答2:

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
}

回答3:

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

@matt has an interesting solution where he defines his own reverse operator and calls it >>>. It doesn't take much code to define and is used like this:
```
for index in 5>>>0 {
    print(index)
}

// 4
// 3
// 2
// 1
// 0
```
Check out On C-Style For Loops Removed from Swift 3

回答4:

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

回答5:

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

回答6:

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
*/

回答7:

In Swift 4 and latter

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

回答8:

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
}

回答9:

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.

回答10:

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

回答11:

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.

回答12:

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

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

回答13:

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
}

回答14:

For this, you can able to use while loop which is very easy

var i = 10
while i > 1{
   print(i)
   i = i - 1
}

But still, if you want to use for loop then use reversed() method

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

回答15:

You can simply do this if you have an array

func reverseArray(a: [Int]) -> [Int] {
   var reverseArr = [Int]()
   for i in stride(from: a.count-1, to: -1, by: -1){
      reverseArr.append(a[i])
   }
   return reverseArr
}

来源：https://stackoverflow.com/questions/24508592/how-to-iterate-for-loop-in-reverse-order-in-swift

标签

ios

swift

swift-playground

How to iterate for loop in reverse order in swift?

问题

回答1:

回答2:

回答3:

reversed: numbers in a range

reversed: elements in SequenceType

reversed: elements with an index

stride: numbers

stride: elements of SequenceType

Notes

回答4:

回答5:

回答6:

#1. Using ClosedRange reversed() method

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

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

#4. Using AnyIterator init(_:) initializer

回答7:

回答8:

回答9:

回答10:

回答11:

回答12:

回答13:

回答14:

回答15:

`reversed`: numbers in a range

`reversed`: elements in `SequenceType`

`reversed`: elements with an index

#1. Using `ClosedRange` `reversed()` method

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

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

#4. Using `AnyIterator` `init(_:)` initializer