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
}