Swift. Get Binary String from an Integer

Get signed integer from swift string of binary

Playing around you can get the desired result as follows:

let binaryString = "1111111110011100"
print(Int(binaryString, radix: 2)!)
print(UInt16(binaryString, radix: 2)!)
print(Int16(bitPattern: UInt16(binaryString, radix: 2)!))

Output:

65436

65436

-100

The desired result comes from creating a signed Int16 using the bit pattern of a UInt16.

Swift. Get binary string from an integer

let binaryString = String(2, radix: 2)
let other = String(count: 8 - binaryString.characters.count, repeatedValue: Character("0")) + binaryString

How I can convert int to binary string swift?

There is a function that can convert to integer Int("000100001010", radix: 2)
To append zeroes, use this:
let str = String(repeating: "0", count: amount)

How to convert a decimal number to binary in Swift?

You can convert the decimal value to a human-readable binary representation using the String initializer that takes a radix parameter:

let num = 22
let str = String(num, radix: 2)
print(str) // prints "10110"

If you wanted to, you could also pad it with any number of zeroes pretty easily as well:

Swift 5

func pad(string : String, toSize: Int) -> String {
  var padded = string
  for _ in 0..<(toSize - string.count) {
    padded = "0" + padded
  }
    return padded
}

let num = 22
let str = String(num, radix: 2)
print(str) // 10110
pad(string: str, toSize: 8)  // 00010110

Swift - Convert a binary string to its ascii values

You may need to split the input binary digits into 8-bit chunks, and then convert each chunk to an ASCII character. I cannot think of a super simple way:

var binaryBits = "010010000110010101111001"

var index = binaryBits.startIndex
var result: String = ""
for _ in 0..<binaryBits.characters.count/8 {
    let nextIndex = binaryBits.index(index, offsetBy: 8)
    let charBits = binaryBits[index..<nextIndex]
    result += String(UnicodeScalar(UInt8(charBits, radix: 2)!))
    index = nextIndex
}
print(result) //->Hey

Convert a binary string into an array of integers in swift 3

A possible solution (now updated for Swift 4 and later):

let zeroOneString = "01101010101001010111"

let resultArray = zeroOneString.compactMap { char in
    char == "0" ? 0 : char == "1" ? 1 : nil
}

print(resultArray) // [0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1]

zeroOneString.compactMap maps the sequences of all characters in the string.
The closure maps "0" to 0 and "1" to 1. Everything else is mapped to nil and ignored by compactMap.

String of Binary Values to Binary File

If you have a string with a series of “0” and “1” characters, and want a binary representation, you can do:

extension String {
    var binaryData: Data {
        let values = try! NSRegularExpression(pattern: "[01]{8}")
            .matches(in: self, range: NSRange(startIndex..., in: self))
            .compactMap { Range($0.range, in: self) }
            .compactMap { self[$0] }
            .compactMap { UInt8($0, radix: 2) }
        return Data(values)
    }
}

Thus

let input = "010000010100001001000011"
let data = input.binaryData               // 0x41 0x42 0x43

You can then write that Data to a file, or do whatever you want with it.

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