String indices in Swift 2

Question

I decided to learn Swift and I decided to start with Swift 2 right away.

So here is a very basic example that's similar to one of the examples from Apple's own e-book about Swift

let greeting = "Guten Tag"

for index in indices(greeting) {
    print(greeting[index])
}

I tried this in the playground of Xcode 7 and I received the following error

Cannot invoke 'indices' with an argument list of type '(String)'

I also tried the same thing with Xcode 6 (which is Swift 1.2 AFAIK) and it worked as expected.

Now, my question is: Is this

An error in Xcode 7, it's still a beta release after all, or
Something that just doesn't work anymore with Swift 2 and the e-book just isn't fully updated yet?

Also: If the answer is "2", how would you replace indices(String) in Swift 2?

Eric Aya · Accepted Answer · 2015-06-11 10:40:30Z

10

In a Playground, if you go to menu View > Debug Area > Show debug area, you can see the full error in the console:

/var/folders/2q/1tmskxd92m94__097w5kgxbr0000gn/T/./lldb/94138/playground29.swift:5:14: error: 'indices' is unavailable: access the 'indices' property on the collection for index in indices(greeting)

Also, Strings do not conform to SequenceTypes anymore, but you can access their elements by calling characters.

So the solution for Swift 2 is to do it like this:

let greeting = "Guten Tag"

for index in greeting.characters.indices {
    print(greeting[index])
}

Result:

G
u
t
e
n

T
a
g

Of course, I assume your example is just to test indices, but otherwise you could just do:

for letter in greeting.characters {
    print(letter)
}

edited Jun 11, 2015 at 10:40

answered Jun 11, 2015 at 10:35

Eric Aya

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2 Comments

mmgross Over a year ago

Thanks. And yes, I was just testing indices and already aware of the easier solution since characters is mentioned earlier in the book. Also your hint about the debugging console was very helpful.

Eric Aya Over a year ago

You're welcome. Indeed, I did hesitate about adding this last comment as I was pretty sure you were aware of it. :) But I added it eventually because I thought it could help future readers to be sure the topic of the post was indices and not actually accessing the characters.

Daniel Klöck · Accepted Answer · 2015-09-24 07:39:50Z

1

Just for completion, I have found a very simple way to get characters and substrings out of strings (this is not my code, but I can't remember where I got it from):

include this String extension in your project:

extension String {

    subscript (i: Int) -> Character {
        return self[self.startIndex.advancedBy(i)]
    }

    subscript (i: Int) -> String {
        return String(self[i] as Character)
    }

    subscript (r: Range<Int>) -> String {
        return substringWithRange(Range(start: startIndex.advancedBy(r.startIndex), end: startIndex.advancedBy(r.endIndex)))
    }
}

this will enable you to do:

print("myTest"[3]) //the result is "e"
print("myTest"[1...3]) //the result is "yTe"

answered Sep 24, 2015 at 7:39

Daniel Klöck

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Comments

Narendra G · Accepted Answer · 2015-09-24 10:44:41Z

-1

Here is the code that you looking:

var middleName :String? = "some thing"
for index in (middleName?.characters.indices)! {
// do some thing with index
}

answered Sep 24, 2015 at 10:44

Narendra G

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