How to write a non-C-like for-loop in Swift 2.2+? - swift

Question

I have updated Xcode (7.3) and there are a lot of changes; C-like for expressions will be deprecated. For a simple example,
for var i = 0; i <= array.count - 1; i++
{
//something with array[i]
}
How do I write this clear and simple C-like for-loop to be compliant with the new changes?
for var i = 0, j = 1; i <= array.count - 2 && j <= array.count - 1; i++, j++
{
//something with array[i] and array[j]
}
Update.
One more variant
for var i = 0; i <= <array.count - 1; i++
{
for var j = i + 1; j <= array.count - 1; j++
{
//something with array[i] and array[j]
}
}
And more ...
for var i = 0, j = 1, g = 2; i <= array.count - 3 && j <= array.count - 2 && g <= array.count - 1; i++, j++, g++
{
//something with array[i] and array[j] and array[g]
}
Update2 After several suggestions for me while loop is preferable universal substitution for all cases more complicated than the simple example of C-like for-loop (suitable for for in expression). No need every time to search for new approach.
For instance: Instead of
for var i = 0; i <= <array.count - 1; i++
{
for var j = i + 1; j <= array.count - 1; j++
{
//something with array[i] and array[j]
}
}
I can use
var i = 0
while i < array.count
{
var j = i + 1
while j < array.count
{
//something with array[i] and array[j]
j += 1
}
i += 1
}

Answer 1

charl's (old) answer will crash. You want 0..<array.count:
for index in 0..<array.count {
// ...
}
If you want something like your i/j loop you can use stride and get i's successor:
for i in 0.stride(through: array.count, by: 1) {
let j = i.successor()
// ...
}
Just make sure to check i.successor() in case you go out of bounds.

Answer 2

for var i = 0; i <= array.count - 1; i++ {
//something with array[i]
}
Here you don't need the element index at all, so you can simply
enumerate the array elements:
for elem in array {
// Do something with elem ...
}
for var i = 0, j = 1; i <= array.count - 2 && j <= array.count - 1; i++, j++ {
//something with array[i] and array[j]
}
To iterate over pairs of adjacent elements, use zip()
and dropFirst():
for (x, y) in zip(array, array.dropFirst()) {
// Do something with x and y ...
print(x, y)
}
Output:
1 2
2 3
3 4
4 5
For other distances, use dropFirst(n):
for (x, y) in zip(array, array.dropFirst(3)) {
// Do something with x and y ...
print(x, y)
}
Output:
1 4
2 5
There are probably many solutions to do
for var i = 0; i <= <array.count - 1; i++ {
for var j = i + 1; j <= array.count - 1; j++ {
//something with array[i] and array[j]
}
}
without a C-style for-loop, here is one:
for (index, x) in array.enumerate() {
for y in array.dropFirst(index + 1) {
print(x, y)
}
}

Answer 3

If you want to do something with subsequent pairs there are many other ways to do it.
Something like this would work...
var previousItem = array.first
for index in 1..<array.count {
let currentItem = array[index]
// do something with current and previous items
previousItem = currentItem
}

Answer 4

for (i, j) in zip(array.dropLast(), array.dropFirst())
{
// something
}
What you're really doing here is enumerating two parallel sequences. So, create those sequences and use zip to turn them into a single sequence.

Answer 5

Do enumeration
let suits = ["♠︎", "♥︎", "♣︎", "♦︎"]
for (i, suite) in suits.enumerate() {
// ...
}
or to compare neighbors
import Foundation
let suits = ["♠︎", "♥︎", "♣︎", "♦︎"]
for (i, suite1) in suits.enumerate() {
let j = i.successor()
if j < suits.count {
let suite2 = suits[j]
// ...
}
}
or zipping and enumerating
let suits = ["♠︎", "♥︎", "♣︎", "♦︎"]
let combination = zip(suits, suits.dropFirst())
for (i, (s1,s2)) in combination.enumerate() {
print("\(i): \(s1) \(s2)")
}
result
0: ♠︎ ♥︎
1: ♥︎ ♣︎
2: ♣︎ ♦︎

Answer 6

Worst case, you can convert it to a while loop.
var i = 0
var j = 1
while i <= array.count -2 && j <= array.count - 1 {
// something
i += 1
j += 1
}
-- EDIT --
Because you said, "while loop is preferable universal substitution for all cases more complicated than the simple example of C-like for-loop"... I feel the need to expand on my answer. I don't want to be responsible for a bunch of bad code...
In most cases, there is a simple for-in loop that can handle the situation:
for item in array {
// do something with item
}
for (item1, item2) in zip(array, array[1 ..< array.count]) {
// do something with item1 and item2
}
for (index, item1) in array.enumerate() {
for item2 in array[index + 1 ..< array.count] {
// do soemthing with item1 and item2
}
}
For your last case, you might be justified using a for look, but that is an extremely rare edge case.
Don't litter your code with for loops.

Answer 7

to compare neighbouring elements from the same array you can use
let arr = [1,2,2,5,2,2,3,3]
arr.reduce(nil) { (i, j)->Int? in
if let i = i {
print(i,"==",j,"is",i == j)
}
return j
}
it prints
1 == 2 is false
2 == 2 is true
2 == 5 is false
5 == 2 is false
2 == 2 is true
2 == 3 is false
3 == 3 is true
more 'generic' approach without using subscript but separate generators
let arr1 = [1,2,3,4,5,6,7,8,9,0]
var g1 = arr1.generate()
var g2 = (arr1.dropFirst(5) as AnySequence).generate()
var g3 = (arr1.dropFirst(6) as AnySequence).generate()
while true {
if let a1 = g1.next(),
let a2 = g2.next(),
let a3 = g3.next() {
print(a1,a2,a3)
} else {
break
}
}
/* prints
1 6 7
2 7 8
3 8 9
4 9 0
*/