Dealing with Octal Numbers in Swift

Dealing with Octal numbers in swift

Using string conversion functions is pretty horrible in my option. How about something like this instead:

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

How to convert from Octal to Decimal in Swift

From Octal to Decimal

There is a specific Int initializer for this

let octal = 10
if let decimal = Int(String(octal), radix: 8) {
    print(decimal) // 8
}

From Decimal to Octal

let decimal = 8
if let octal = Int(String(decimal, radix: 8)) {
    print(octal) // 10
}

Note 1: Please pay attention: parenthesis are different in the 2 code snippets.

Note 2: Int initializer can fail for string representations of number with more exotic radixes. Please read the comment by @AMomchilov below.

Only allowing octal numbers

I would replace all the scanf and harcoded functions with fgets and strtol:

enum {oct = 8, dec = 10, hex = 16};

int to_number(int base, bool *match)
{
    char str[32];
    long num = 0;

    *match = false;
    if (fgets(str, sizeof str, stdin))
    {
        char *ptr;

        num = strtol(str, &ptr, base);
        *match = (*ptr == '\n'); // match is true if strtol stops scanning
                                 // at the newline left by fgets
    }
    return (int)num;
}

...

bool match;
int num = to_number(oct, &match);

printf("%d %s an octal\n", num, match ? "is" : "is not");

EDIT:

As pointed out by @klutt in comments, since strtol returns a long, ideally the function should also return long in order to avoid the cast, then, if you really need an int, you can cast the return of the function:

long to_number(int base, bool *match);

int num = (int)to_number(oct, &match);

How to convert a fractional Octal to Hex fraction and vice versa in Swift 3?

Convert hex to binary, pad to make binary string a multiple of 3 in length, then convert binary to octal:

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"

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And the other direction (octal -> hexadecimal):

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"

Octal Numbering System

Easy. Just continue the pattern from the left side of the dot:

1*8^1 + 0*8^0 + 3*8^-1 + 4*8^-2 = 8.4375

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