如何在 Swift 3 中将八进制分数转换为十六进制分数，反之亦然？

Question

我正在 Swift 中寻找一种将八进制小数转换为十六进制小数的方法，反之亦然 3. 无需第三方库和 Foundation 的自行编写的代码。例如，此代码将八进制整数转换为十六进制：

for num in OctalVal {     
            switch num {
            case "0": result = result * 8
            case "1": result = result * 8 + 1
            case "2": result = result * 8 + 2
            case "3": result = result * 8 + 3
            case "4": result = result * 8 + 4
            case "5": result = result * 8 + 5
            case "6": result = result * 8 + 6
            case "7": result = result * 8 + 7
            default: return "Error"    
            } 
        }
        n2 = result
        while n2 > 0 {
            c2 = n2 % 16
            n2 = n2 / 16
            k2.append("\(c2)")
        }

        for i in k2.reversed() {
            if(i == "10"){
                fs2 += "A"
            }
            else if(i == "11") {
                fs2 += "B"
            }
            else if(i == "12") {
                fs2 += "C"
            }
            else if(i == "13") {
                fs2 += "D"
            }
            else if(i == "14") {
                fs2 += "E"
            }
            else if(i == "15") {
                fs2 += "F"
            }
            else {
                fs2 += "\(i)"
            }  
        }   
        return "\(fs2)"  
    }

而这一个十六进制到八进制：

for num in Left {
            switch num {
            case "0": result += "0000"
            break;
            case "1": result += "0001"
            break;
            case "2": result += "0010"
            break;
            case "3": result += "0011"
            break;
            case "4": result += "0100"
            break;
            case "5": result += "0101"
            break;
            case "6": result += "0110"
            break;
            case "7": result += "0111"
            break;
            case "8": result += "1000"
            break;
            case "9": result += "1001"
            break;
            case "A": result += "1010"
            break;
            case "B": result += "1011"
            break;
            case "C": result += "1100"
            break;
            case "D": result += "1101"
            break;
            case "E": result += "1110"
            break;
            case "F": result += "1111"
            break;
            default: return "Error"
            }
        }

例如：12D.3AF9F 到 455.1657476（八进制）

关于小数部分有什么想法吗？感谢您关注我的请求

Answer 1

将十六进制转换为二进制，填充使二进制字符串的长度成为 3 的倍数，然后将二进制转换为八进制：

func hexadecimalFractToOctal(_ hex: String) -> String {
    let hexToBin = ["0": "0000", "1": "0001", "2": "0010", "3": "0011",
                    "4": "0100", "5": "0101", "6": "0110", "7": "0111",
                    "8": "1000", "9": "1001", "A": "1010", "B": "1011",
                    "C": "1100", "D": "1101", "E": "1110", "F": "1111"]

    let binToOct = ["000": "0", "001": "1", "010": "2", "011": "3",
                    "100": "4", "101": "5", "110": "6", "111": "7"]

    // Convert hex string to binary
    var bin = ""
    for char in hex.characters {
        bin += hexToBin[String(char)] ?? ""
    }

    // Pad the string to a multiple of 3 binary digits
    bin += ["", "00", "0"][bin.characters.count % 3]

    var binChars = bin.characters

    var oct = ""

    // Convert binary string to octal 3 digits at a time
    while binChars.count > 0 {
        let b = String(binChars.prefix(3))
        binChars = binChars.dropFirst(3)
        oct += binToOct[b] ?? ""
    }

    return oct
}

print(hexadecimalFractToOctal("3AF9F"))  // "1657476"

而另一个方向（八进制 -> 十六进制）：

func octalFractToHexadecimal(_ oct: String) -> String {
    let octToBin = ["0": "000", "1": "001", "2": "010", "3": "011",
                    "4": "100", "5": "101", "6": "110", "7": "111"]

    let binToHex = ["0000": "0", "0001": "1", "0010": "2", "0011": "3",
                    "0100": "4", "0101": "5", "0110": "6", "0111": "7",
                    "1000": "8", "1001": "9", "1010": "A", "1011": "B",
                    "1100": "C", "1101": "D", "1110": "E", "1111": "F"]

    // Convert octal string to binary
    var bin = ""
    for char in oct.characters {
        bin += octToBin[String(char)] ?? ""
    }

    // Pad the string to a multiple of 4 binary digits
    bin += ["", "000", "00", "0"][bin.characters.count % 4]

    var binChars = bin.characters

    var hex = ""

    // Convert binary string to hexadecimal 4 digits at a time
    while binChars.count > 0 {
        let b = String(binChars.prefix(4))
        binChars = binChars.dropFirst(4)
        hex += binToHex[b] ?? ""
    }

    return hex
}

print(octalFractToHexadecimal("1657476")) // "3AF9F0"

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