CODEWITHSUNDEEP
swiftswift2
Bartłomiej Semańczyk
Bartłomiej Semańczyk

Reputation: 61774

Cannot invoke `join` with an argument list of type (String, [String]) in Swift 2.0

var specializationTitles = ["a", "b", "c", "d"]
let outputString = join(" / ", specializationTitles)

got an error:

Cannot invoke join with an argument list of type (String, [String])

How to solve this?

Upvotes: 15

Views: 5185

Answers (2)

Bartłomiej Semańczyk
Bartłomiej Semańczyk

Reputation: 61774

let separator = " / "
let outputString = separator.join(specializationTitles)

With Xcode7beta6:

specializationTitles.joinWithString(" / ")

With Xcode7 release version:

specializationTitles.joinWithSeparator(" / ")

Upvotes: 21

Nagendra Rao
Nagendra Rao

Reputation: 7152

In Swift 2.0 ( which is default on Xcode 7), you will have to use, joinWithSeparator

specializationTitles.joinWithSeparator(" / ")

Upvotes: 2

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