How to iterate for loop in reverse order in swift

Question

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

User · Answer

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

User · Answer

You can consider using the C-Style while loop instead  This works just fine in Swift 3   var i   5  while i  gt  0        print i      i -  1

User · Answer

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   - gt  ReversedCollection lt ClosedRange lt Bound gt  gt       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    lt  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 lt T gt  first  T  next   escaping  T  - gt  T   - gt  UnfoldFirstSequence lt T gt       Returns a sequence formed from first and repeated lazy applications of next    Usage   let unfoldSequence   sequence first  5  next         0  gt  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 lt T gt  from start  T  through end  T  by stride  T Stride  - gt  StrideThrough lt T gt  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    - gt  AnyIterator lt Element gt  Element        Creates an iterator that wraps the given closure in its next   method    Usage   var index   5  guard index  gt   0 else   fatalError  index must be positive or equal to zero      let iterator   AnyIterator      - gt  Int  in     defer   index   index - 1       return index  gt   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  - gt  AnyIterator lt Int gt            var index   self         guard index  gt   endIndex else   fatalError  self must be greater than or equal to endIndex              let iterator   AnyIterator      - gt  Int  in             defer   index   index - 1               return index  gt   endIndex   index   nil                   return iterator           let iterator   5 iterateDownTo 0   for index in iterator       print index        Prints  5 4 3 2 1 0

User · Answer

var sum1   0 for i in 0   100      sum1    i   print  sum1   for i in  10   100  reverse        sum1    i   print sum1

User · Answer

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

User · Answer

In Swift 4 and latter      let count   50  For example     for i in  1   count  reversed             print i

User · Answer

For me  this is the best way   var arrayOfNums    1 4 5 68 9 10   for i in 0   lt arrayOfNums count       print arrayOfNums arrayOfNums count - i - 1

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

For Swift 2 0 and above you should apply reverse on a range collection  for i in  0    lt  10  reverse          process     It has been renamed to  reversed   in Swift 3 0

User · Answer

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

User · Answer

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   lt 5       print index        0    1    2    3    4   Backward  for index in  0   lt 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   lt 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  gt  gt  gt   It doesn t take much code to define and is used like this   for index in 5 gt  gt  gt 0       print index        4    3    2    1    0  Check out On C-Style For Loops Removed from Swift 3

User · Answer

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   lt 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  gt  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 lt  Range lt Int gt   gt   so this works fine   var count   0 for i in  1   1 000 000 000 000  reverse         count    1     if count  gt  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

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

Examples related to ios

Examples related to swift

Examples related to swift-playground