Cannot invoke 'split' with an argument list of type '(String, (String) -> Bool)' in SWIFT 2

Question

The error: Cannot invoke 'split' with an argument list of type '(String, (String) -> Bool)'

Code i want to use: let nameArr = split(name) {$0 == "."}

Have a look at (the Swift 2 updates to) stackoverflow.com/a/25229901/1187415. — Martin R
– Martin R, Commented Jul 30, 2015 at 15:05

Vkt0r · Accepted Answer · 2015-07-30 15:13:29Z

4

Strings are No Longer Collections, String no longer conforms to CollectionType. You can use other alternatives like the function componentsSeparatedByString:

var name = "Victor.Hello.GYTT" 
let nameArr = name.componentsSeparatedByString(".") // [Victor, Hello, GYTT]

Another option is using the characters property:

let nameArr = split(name.characters) { $0 == "." }.map { String($0) }

With the new .init syntax in Xcode 7 beta 2, where init "can now be referenced like static methods" like in the following way:

let nameArr = split(name.characters) { $0 == "." }.map { String.init }

Or making the String conforms the protocol too, but Apple made a decision to remove String conformance to Sliceable, be carefull.

You can read more about notables changes in Changes to the Swift Standard Library in 2.0 beta 1 in the blog of @AirSpeedVelocity. Really nice.

I hope this help you.

edited Jul 30, 2015 at 15:13

answered Jul 30, 2015 at 15:07

Vkt0r

23.5k15 gold badges86 silver badges103 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Omar Shiltawi Over a year ago

am getting an error Initializer for conditional binding must have Optional type, not 'NSURL'

Vkt0r Over a year ago

Provide the declaration of your variable name, see my answer

Omar Shiltawi Over a year ago

am really new to swift so idk what to do here is the error Image

Vkt0r Over a year ago

I saw you error , but what's the problem? you can't use the split in this way anymore , you need to change your line let nameArr = split(name) {$0 == "."} for any of the above I explained you.

Omar Shiltawi Over a year ago

sorry for being a pain but neither of them worked for me each one shows an error in a different way like for example look at this image

|

Collectives™ on Stack Overflow

Cannot invoke 'split' with an argument list of type '(String, (String) -> Bool)' in SWIFT 2

1 Answer 1

6 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related