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+--- Thread: How to iterate for loop in reverse order in swift? (/Thread-How-to-iterate-for-loop-in-reverse-order-in-swift)
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How to iterate for loop in reverse order in swift? - preregisters943240 - 07-18-2023
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][1]
[1]:
RE: How to iterate for loop in reverse order in swift? - Sirennav - 07-18-2023
This will works decrement by one in reverse order.
let num1 = [1,2,3,4,5]
for item in nums1.enumerated().reversed() {
print(item.offset) // Print the index number: 4,3,2,1,0
print(item.element) // Print the value :5,4,3,2,1
}
Or you can use this index, value property
let num1 = [1,2,3,4,5]
for (index,item) in nums1.enumerated().reversed() {
print(index) // Print the index number: 4,3,2,1,0
print(item) // Print the value :5,4,3,2,1
}
RE: How to iterate for loop in reverse order in swift? - impair375 - 07-18-2023
Reversing an array can be done with just single step **.reverse()**
var arrOfnum = [1,2,3,4,5,6]
arrOfnum.reverse()
RE: How to iterate for loop in reverse order in swift? - japanization599302 - 07-18-2023
> 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])
}
RE: How to iterate for loop in reverse order in swift? - regimen811 - 07-18-2023
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()`][1]. `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()`][2] 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:)`][3]. `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:)`][4]. `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:)`][5]:
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(_:)`][6]. `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
*/
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RE: How to iterate for loop in reverse order in swift? - jew86 - 07-18-2023
*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][2], 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][3] 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][1]
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RE: How to iterate for loop in reverse order in swift? - chickie810 - 07-18-2023
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.
RE: How to iterate for loop in reverse order in swift? - scissorbill583137 - 07-18-2023
**In Swift 4 and latter**
let count = 50//For example
for i in (1...count).reversed() {
print(i)
}
RE: How to iterate for loop in reverse order in swift? - glynxlayk - 07-18-2023
**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
RE: How to iterate for loop in reverse order in swift? - aminat - 07-18-2023
**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
}