无法推断通用参数'Self'(快速4)

时间:2018-01-05 11:19:19

标签: xcode swift4

我想为字符串创建一个子字符串扩展名,我尝试了以下2种方法但不成功:

extension String
{
    func substringToFirstChar(of char: Character) -> String
    {
        let pos = self.range(of: String(char))
        let subString = self[..<pos?.lowerBound]   
        return String(subString)
    }
}

extension String
{
    func substringToFirstChar(of char: Character) -> String
    {
        let pos = self.index(of: char)
        let subString = self[..<pos]
        return String(subString)
    }
}
  

xcode提示错误:无法推断让subString = self [.. “行<。>

怎么做?

1 个答案:

答案 0 :(得分:9)

range(of / index(of返回一个可选项。您必须在范围表达式

中打开可选项
extension String
{
    func substringToFirstChar(of char: Character) -> String?
    {
        guard let pos = self.range(of: String(char))?.lowerBound else { return nil }
        // or  guard let pos = self.index(of: char) else { return nil }
        let subString = self[..<pos]
        return String(subString)
    }
}

或者 - 避免可选 - 如果没有匹配则返回未更改的字符串

extension String
{
    func substringToFirstChar(of char: Character) -> String
    {
        guard let pos = self.range(of: String(char))?.lowerBound else { return self }
        // or  guard let pos = self.index(of: char) else { return self }
        let subString = self[..<pos]
        return String(subString)
    }
}