加入joinWithSeparator为int数组？

Question

在新的Swift 2样式连接必须由joinWithSeparator替换。但我得到了错误消息，发现这个模棱两可的参考：

var distribCharactersInt = [Int](count:lastIndex + 1, repeatedValue:0) 
...       
let DistributionCharacterString = distribCharactersInt.joinWithSeparator(",") 

我忘了什么？

Answer 1

有两种joinWithSeparator()方法。一个需要序列的 序列：

extension SequenceType where Generator.Element : SequenceType { 
    /// Returns a view, whose elements are the result of interposing a given 
    /// `separator` between the elements of the sequence `self`. 
    /// 
    /// For example, 
    /// `[[1, 2, 3], [4, 5, 6], [7, 8, 9]].joinWithSeparator([-1, -2])` 
    /// yields `[1, 2, 3, -1, -2, 4, 5, 6, -1, -2, 7, 8, 9]`. 
    @warn_unused_result 
    public func joinWithSeparator<Separator : SequenceType where Separator.Generator.Element == Generator.Element.Generator.Element>(separator: Separator) -> JoinSequence<Self> 
} 

和其他需要串序列（和字符串作为分隔符）：

extension SequenceType where Generator.Element == String { 
    /// Interpose the `separator` between elements of `self`, then concatenate 
    /// the result. For example: 
    /// 
    ///  ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz" 
    @warn_unused_result 
    public func joinWithSeparator(separator: String) -> String 
} 

你可能会想要使用第二种方法，但那么你有 将数字转换为字符串：

let distribCharactersInt = [Int](count:5, repeatedValue:0) 
let distributionCharacterString = distribCharactersInt.map(String.init).joinWithSeparator(",") 
print(distributionCharacterString) // 0,0,0,0,0