スウィフト3では小数点以下の八進数を16進数に変換し、逆も同様です。

Question

Swift 3では小数点以下の8進数を16進数に変換する方法を探しています。第三者のライブラリとFoundationを使用しない自己記述コード。

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)" 
    } 

そして、この1ヘックスのオクタルへ：

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：小数に関する12D.3AF9F

（8進数）455.1657476への任意のアイデアをたとえば、このコードは16進数8進数整数に変換します部？ はその後進バイナリ変換し、バイナリ文字列の長さの3の倍数にするバイナリ、パッドに私の要求

Answer 1

変換の六角にご清聴ありがとうございました：

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" 

---

そして、他の方向（16進数→16進数）：

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"