问题

I am trying to do a very simple piece of code in Swift playgrounds.

var word = \"Zebra\"

for i in word {
  print(i)
}

However, I always get an error on line 3.

\'String\' does not have a member named \'Generator\'

Any ideas on why this doesn\'t work? Note: I am working in Xcode 7, with Swift 2.0 (Strings and Characters).

回答1:

As of Swift 2, String doesn't conform to SequenceType. However, you can use the characters property on String. characters returns a String.CharacterView which conforms to SequenceType and so can be iterated through with a for loop:

let word = "Zebra"

for i in word.characters {
    print(i)
}

Alternatively, you could add an extension to String to make it conform to SequenceType:

extension String: SequenceType {}

// Now you can use String in for loop again.
for i in "Zebra" {
    print(i)
}

Although, I'm sure Apple had a reason for removing String's conformance to SequenceType and so the first option seems like the better choice. It's interesting to explore what's possible though.

回答2:

String doesn't conform to SequenceType anymore. However you can access the characters property of it this way:

var word = "Zebra"

for i in word.characters {
    print(i)
}

Note that the documentation hasn't been updated yet.

回答3:

Swift 4

Forin loop:

let word = "Swift 4"
for i in word {
    print(i)
}

map example:

let word = "Swift 4"
_ = word.map({ print($0) })

forEach example:

let word = "Swift 4"
word.forEach({ print($0) })

回答4:

Swift 3.0.1

Use the indices property of the characters property to access all of the indices of individual characters in a string.

let greeting = "Guten Tag!"
for index in greeting.characters.indices {
print("\(greeting[index]) ", terminator: "")
}
// Prints "G u t e n   T a g ! "

visit https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/StringsAndCharacters.html

来源：https://stackoverflow.com/questions/30767594/iterate-through-a-string-swift-2-0

标签

swift