在此下方，您可以看到我的代码。我注释了导致错误的行。错误消息：“下标调用中没有完全匹配”。您知道如何避免此错误吗？感谢您的帮助！
let dic = ["a": 1, "b": 2, "c": 3, "d": 4, "e": 5, "f": 6, "g": 7, "h": 8, "i": 9, "j": 10, "k": 11, "l": 12, "m": 13, "n": 14, "o": 15, "p": 16, "q": 17, "r": 18, "s": 19, "t": 20, "u": 21, "v": 22, "w": 23, "x": 24, "y": 25, "z": 26]
var newwrd = ""
for var i in str ?? "" {
let ci = dic[i] // This line causes the error
}
其实你应该得到错误
无法使用类型为'String.Element'（又名'字符'）的参数对类型为[[String：Int]'的值进行下标
str显然String?，枚举字符串时的元素类型为，Character但订阅类型必须为String。
let dic = ["a": 1, "b": 2, "c": 3, "d": 4, "e": 5, "f": 6, "g": 7, "h": 8, "i": 9, "j": 10, "k": 11, "l": 12, "m": 13, "n": 14, "o": 15, "p": 16, "q": 17, "r": 18, "s": 19, "t": 20, "u": 21, "v": 22, "w": 23, "x": 24, "y": 25, "z": 26]
for i in str ?? "" { // no need for var i
let ci = dic[String(i)] ?? 0
print(ci)
}
如果字符串包含不在字典中的字符，则结果为0。
没有助手字典有更短的方法
for i in str ?? "" where ("a"..."z") ~= i {
let ci = Int(i.asciiValue!) - 96
print(ci)
}
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