How to convert a decimal number to binary in Swift

Question

How can I convert Int to UInt8 in Swift? Example. I want to convert number 22 to 0b00010110

var decimal = 22
var binary:UInt8 = ??? //What should I write here?

Answer 1

Swift 5.1 / Xcode 11

Thanks Gustavo Seidler. My version of his solution is complemented by spaces for readability.

extension BinaryInteger {
    var binaryDescription: String {
        var binaryString = ""
        var internalNumber = self
        var counter = 0

        for _ in (1...self.bitWidth) {
            binaryString.insert(contentsOf: "\(internalNumber & 1)", at: binaryString.startIndex)
            internalNumber >>= 1
            counter += 1
            if counter % 4 == 0 {
                binaryString.insert(contentsOf: " ", at: binaryString.startIndex)
            }
        }

        return binaryString
    }
}

Examples:

UInt8(9).binaryDescription      // "0000 1001"
Int8(5).binaryDescription       // "0000 0101"
UInt16(1945).binaryDescription  // "0000 0111 1001 1001"

Int16(14).binaryDescription     // "0000 0000 0000 1110"
Int32(6).binaryDescription      // "0000 0000 0000 0000 0000 0000 0000 0110"
UInt32(2018).binaryDescription  // "0000 0000 0000 0000 0000 0111 1110 0010"

Answer 2

If you want binary to have the value of 22, just assign it that: binary = 22 or you could write it as binary = 0b00010110; the two statements are equivalent.

Answer 3

swift 4.1

extension String {
    public func pad(with padding: Character, toLength length: Int) -> String {
        let paddingWidth = length - self.count
        guard 0 < paddingWidth else { return self }

        return String(repeating: padding, count: paddingWidth) + self
    }
}


extension UInt8 {
     public func toBits() -> String
     {
          let a = String( self, radix : 2 )
          let b = a.pad(with: "0", toLength: 8)
          return b
     }
}

func showBits( _ list: [UInt8] )
{
    for num in list
    {
        showBits(num)
    }
}

func showBits( _ num: UInt8 )
{
    //print(num, String( num, radix : 2 ))
    print( "\(num) \t" +   num.toBits())
}

let initialBits :UInt8 = 0b00001111
let invertedBits = ~initialBits
showBits( [initialBits, invertedBits] )

result

15 00001111

240 11110000

good for you~

Answer 4

Here's how I would do it:

extension String {
    public func pad(with padding: Character, toLength length: Int) -> String {
        let paddingWidth = length - self.characters.count
        guard 0 < paddingWidth else { return self }

        return String(repeating: padding, count: paddingWidth) + self
    }
}

String(0b1010, radix: 2).pad(with: "0", toLength: 8) //00001010

Answer 5

Since none of the solutions contemplate negative numbers, I came up with a simple solution that basically reads the number's internal representation and pads it automatically to the width of its type. This should work on all BinaryInteger types.

extension BinaryInteger {
    var binaryDescription: String {
        var binaryString = ""
        var internalNumber = self
        for _ in (1...self.bitWidth) {
            binaryString.insert(contentsOf: "\(internalNumber & 1)", at: binaryString.startIndex)
            internalNumber >>= 1
        }
        return "0b" + binaryString
    }
}

Examples:

UInt8(22).binaryDescription     // "0b00010110"
Int8(60).binaryDescription      // "0b00111100"
Int8(-60).binaryDescription     // "0b11000100"
Int16(255).binaryDescription    // "0b0000000011111111"
Int16(-255).binaryDescription   // "0b1111111100000001"

Answer 6

So I had this come up recently. The other generic solutions didn't work for me, due to various issues. Anyway, here's my solution (Swift 4):

extension String {
  init<B: FixedWidthInteger>(fullBinary value: B) {
    self = value.words.reduce(into: "") {
      $0.append(contentsOf: repeatElement("0", count: $1.leadingZeroBitCount))
      $0.append(String($1, radix: 2))
    }
  }
}

Tests:

// result: 0000000000000000000000000000000000000000000000000000000000001001
String(fullBinary: 9)
// result: 1111111111111111111111111111111111111111111111111111111100000000
String(fullBinary: -256)

// result: 1111111111111111111111111111111111111111111111111101100011110001
String(fullBinary: -9999)
// result: 0000000000000000000000000000000000000000000000000010011100001111
String(fullBinary: 9999)

// result: 1100011000000000000000000000000000000000000011110110100110110101
String(fullBinary: 14267403619510741429 as UInt)