Swift 3.0 iterate over String.Index range - swift

Question

The following was possible with Swift 2.2:
let m = "alpha"
for i in m.startIndex..<m.endIndex {
print(m[i])
}
a
l
p
h
a
With 3.0, we get the following error:
Type 'Range' (aka 'Range') does not conform to protocol 'Sequence'
I am trying to do a very simple operation with strings in swift -- simply traverse through the first half of the string (or a more generic problem: traverse through a range of a string).
I can do the following:
let s = "string"
var midIndex = s.index(s.startIndex, offsetBy: s.characters.count/2)
let r = Range(s.startIndex..<midIndex)
print(s[r])
But here I'm not really traversing the string. So the question is: how do I traverse through a range of a given string. Like:
for i in Range(s.startIndex..<s.midIndex) {
print(s[i])
}

Answer 1

You can traverse a string by using indices property of the characters property like this:
let letters = "string"
let middle = letters.index(letters.startIndex, offsetBy: letters.characters.count / 2)
for index in letters.characters.indices {
// to traverse to half the length of string
if index == middle { break } // s, t, r
print(letters[index]) // s, t, r, i, n, g
}
From the documentation in section Strings and Characters - Counting Characters:
Extended grapheme clusters can be composed of one or more Unicode scalars. This means that different characters—and different representations of the same character—can require different amounts of memory to store. Because of this, characters in Swift do not each take up the same amount of memory within a string’s representation. As a result, the number of characters in a string cannot be calculated without iterating through the string to determine its extended grapheme cluster boundaries.
emphasis is my own.
This will not work:
let secondChar = letters[1]
// error: subscript is unavailable, cannot subscript String with an Int

Answer 2

Another option is to use enumerated() e.g:
let string = "Hello World"
for (index, char) in string.characters.enumerated() {
print(char)
}
or for Swift 4 just use
let string = "Hello World"
for (index, char) in string.enumerated() {
print(char)
}

Answer 3

Use the following:
for i in s.characters.indices[s.startIndex..<s.endIndex] {
print(s[i])
}
Taken from Migrating to Swift 2.3 or Swift 3 from Swift 2.2

Answer 4

Iterating over characters in a string is cleaner in Swift 4:
let myString = "Hello World"
for char in myString {
print(char)
}

Answer 5

If you want to traverse over the characters of a String, then instead of explicitly accessing the indices of the String, you could simply work with the CharacterView of the String, which conforms to CollectionType, allowing you access to neat subsequencing methods such as prefix(_:) and so on.
/* traverse the characters of your string instance,
up to middle character of the string, where "middle"
will be rounded down for strings of an odd amount of
characters (e.g. 5 characters -> travers through 2) */
let m = "alpha"
for ch in m.characters.prefix(m.characters.count/2) {
print(ch, ch.dynamicType)
} /* a Character
l Character */
/* round odd division up instead */
for ch in m.characters.prefix((m.characters.count+1)/2) {
print(ch, ch.dynamicType)
} /* a Character
l Character
p Character */
If you'd like to treat the characters within the loop as strings, simply use String(ch) above.
With regard to your comment below: if you'd like to access a range of the CharacterView, you could easily implement your own extension of CollectionType (specified for when Generator.Element is Character) making use of both prefix(_:) and suffix(_:) to yield a sub-collection given e.g. a half-open (from..<to) range
/* for values to >= count, prefixed CharacterView will be suffixed until its end */
extension CollectionType where Generator.Element == Character {
func inHalfOpenRange(from: Int, to: Int) -> Self {
guard case let to = min(to, underestimateCount()) where from <= to else {
return self.prefix(0) as! Self
}
return self.prefix(to).suffix(to-from) as! Self
}
}
/* example */
let m = "0123456789"
for ch in m.characters.inHalfOpenRange(4, to: 8) {
print(ch) /* \ */
} /* 4 a (sub-collection) CharacterView
5
6
7 */

Answer 6

The best way to do this is :-
let name = "nick" // The String which we want to print.
for i in 0..<name.count
{
// Operation name[i] is not allowed in Swift, an alternative is
let index = name.index[name.startIndex, offsetBy: i]
print(name[index])
}
for more details visit here

Answer 7

Swift 4.2
Simply:
let m = "alpha"
for i in m.indices {
print(m[i])
}

Answer 8

Swift 4:
let mi: String = "hello how are you?"
for i in mi {
print(i)
}

Answer 9

To concretely demonstrate how to traverse through a range in a string in Swift 4, we can use the where filter in a for loop to filter its execution to the specified range:
func iterateStringByRange(_ sentence: String, from: Int, to: Int) {
let startIndex = sentence.index(sentence.startIndex, offsetBy: from)
let endIndex = sentence.index(sentence.startIndex, offsetBy: to)
for position in sentence.indices where (position >= startIndex && position < endIndex) {
let char = sentence[position]
print(char)
}
}
iterateStringByRange("string", from: 1, to: 3) will print t, r and i

Answer 10

When iterating over the indices of characters in a string, you seldom only need the index. You probably also need the character at the given index. As specified by Paulo (updated for Swift 4+), string.indices will give you the indices of the characters. zip can be used to combine index and character:
let string = "string"
// Define the range to conform to your needs
let range = string.startIndex..<string.index(string.startIndex, offsetBy: string.count / 2)
let substring = string[range]
// If the range is in the type "first x characters", like until the middle, you can use:
// let substring = string.prefix(string.count / 2)
for (index, char) in zip(substring.indices, substring) {
// index is the index in the substring
print(char)
}
Note that using enumerated() will produce a pair of index and character, but the index is not the index of the character in the string. It is the index in the enumeration, which can be different.