Swift：String.join()如何工作自定义类型？

var a = [1,".join(a.map { $0.description })

或添加此扩展

extension String {
    func join<S : SequenceType where S.Generator.Element : Printable>(elements: S) -> String {
        return self.join(map(elements){ $0.description })
    }

  // use this if you don't want it constrain to Printable
  //func join<S : SequenceType>(elements: S) -> String {
  //    return self.join(map(elements){ "\($0)" })
  //}
}

var a = [1,".join(a)  // works with new overload of join

连接被定义为

extension String {
    func join<S : SequenceType where String == String>(elements: S) -> String
}

这意味着它需要一个字符串序列,你不能传递一个int序列.