Swift 错误：无法将“字符”类型的值转换为预期的参数类型“Unicode.Scalar”

Question

我是 Swift 的新手，并且在使用下面的这个函数时遇到了问题。 我正在使用 SWIFT 5/Xcode 11.3。 该函数旨在删除元音之前的任何字母。 例如，“Brian”将返回“ian”，“Bill”将返回“ill”等。我在下面的 2 行中收到错误消息。

import Foundation

func shortNameFromName(_ name: String) -> String {

    // Definition of characters
    let vowels = CharacterSet(charactersIn: "aeiou")
    var shortName = ""
    let start = name.startIndex

    // Loop through each character in name
    for number in 0..<name.count {

        // If the character is a vowel, remove all characters before current index position
        if vowels.contains(name[name.index(start, offsetBy: number)]) == true { //**ERROR: Cannot convert value of type 'Character' to expected argument type 'Unicode.Scalar'**
            var shortName = name.remove(at: shortName.index(before: shortName.number)) //**ERROR: Cannot use mutating member on immutable value: 'name' is a 'let' constant**
        }
    }

    //convert returned value to lowercase
    return name.lowercased()
}

var str = "Brian" // Expected result is "ian"

shortNameFromName(str)

Answer 1

您可以使用集合的方法func drop(while predicate: (Character) throws -> Bool) rethrows -> Substring而字符串“aeiou”不包含字符并返回一个子字符串：

func shortName(from name: String) -> String { name.drop{ !"aeiou".contains($0) }.lowercased() }

shortName(from: "Brian")  // "ian"    
shortName(from: "Bill")   // "ill"

关于代码中的问题，请通过下面的代码检查注释：

func shortName(from name: String) -> String {
    // you can use a string instead of a CharacterSet to fix your first error
    let vowels =  "aeiou"
    // to fix your second error you can create a variable from your first parameter name
    var name = name
    // you can iterate through each character using `for character in name`
    for character in name {
        // check if the string with the vowels contain the current character
        if vowels.contains(character) {
            // and remove the first character from your name using `removeFirst` method
            name.removeFirst()
        }
    }
    // return the resulting name lowercased
    return name.lowercased()
} 

shortName(from: "Brian")  // "ian"