Swift 3.0 iterate over String.Index range

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)
}

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

Iterating over characters in a string is cleaner in Swift 4:

let myString = "Hello World"    
for char in myString {
    print(char)
}

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 */

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])
}

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Swift 4.2

Simply:

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

Swift 4:

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

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

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.