处理 swift 中的八进制数

Dealing with Octal numbers in swift

我可以轻松地将十进制数转换为八进制数,但我正在尝试相反的操作,但遇到了困难。

let decimal = 11_224_393
let octString = String(rawAddress, radix: 8, uppercase: false)
let octal = octString.toInt()

问题

我想要一个给定八进制数字 Int 的函数,将其作为八进制读入并将其转换为十进制

如:

// oct2dec(777) = 511
// oct2dec(10) = 8

func oct2dec(octal : Int) -> Int {
    // what goes here?
}

在我看来,使用字符串转换函数非常糟糕。换成这样怎么样:

func octalToDecimal(var octal: Int) -> Int {
    var decimal = 0, i = 0
    while octal != 0 {
        var remainder = octal % 10
        octal /= 10
        decimal += remainder * Int(pow(8, Double(i++)))
    }
    return decimal
}

var decimal = octalToDecimal(777) // decimal is 511

只用Swift原生octal literals and initializers (String from integer | Int from String).

let octalInt = 0o1234 // 668
let octalString = "1234" // "1234"

let decimalIntFromOctalString = Int(octalString, radix: 0o10) // 668
let octalStringFromInt = String(octalInt, radix: 0o10) // "1234"

对于您的特定用例:

let decimal = 11_224_393
let octString = String(rawAddress, radix: 0o10, uppercase: false)
guard let octal = Int(octString, radix: 0o10) else {
    print("octString was not a valid octal string:", octString)
}