Question

在新的Swift 2样式中，join必须由joinWithSeparator替换。但是我得到了错误消息，发现了模糊的引用：

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。一个需要一个序列序列：

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>
}

，另一个采用一系列字符串（和一个字符串作为分隔符）：

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
}

你可能想要使用第二种方法，但是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

为了int数组加入joinWithSeparator？

1 个答案: