Base62编码的范围运算符

Question

对于base 62编码，我需要所有62个字母数字字符。 F＃范围运算符为此提供了一个很好的简写。

let alphaNumericCharacters =
    seq {
        yield! [|'a'..'z'|]
        yield! [|'A'..'Z'|]
        yield! [|'0'..'9'|]
    } |> Array.ofSeq

这很简洁，但我很贪心。有没有办法在一行中做到这一点？

Answer 1

let alphaNumericCharacters = Array.concat [| [|'0'..'9'|]; [|'A'..'Z'|]; [|'a'..'z'|] |]

Answer 2

let alphaNumericCharacters = ['a'..'z'] @ ['A'..'Z'] @ ['0'..'9'] |> List.toArray

Answer 3

如果你觉得好笑：

let alphaNumericCharacters = [|Char.MinValue..Char.MaxValue|] |> Array.filter Char.IsLetterOrDigit