Getting the decimal part of a double in Swift

Without converting it to a string, you can round up to a number of decimal places like this:

let x:Double = 1234.5678
let numberOfPlaces:Double = 4.0
let powerOfTen:Double = pow(10.0, numberOfPlaces)
let targetedDecimalPlaces:Double = round((x % 1.0) * powerOfTen) / powerOfTen

Your output would be

0.5678

Same approach as Alessandro Ornano implemented as an instance property of FloatingPoint protocol:

Xcode 11 • Swift 5.1

import Foundation

extension FloatingPoint {
    var whole: Self { modf(self).0 }
    var fraction: Self { modf(self).1 }
}

---

1.2.whole    // 1
1.2.fraction // 0.2

---

If you need the fraction digits and preserve its precision digits you would need to use Swift Decimal type and initialize it with a String:

extension Decimal {
    func rounded(_ roundingMode: NSDecimalNumber.RoundingMode = .plain) -> Decimal {
        var result = Decimal()
        var number = self
        NSDecimalRound(&result, &number, 0, roundingMode)
        return result
    }
    var whole: Decimal { rounded(sign == .minus ? .up : .down) }
    var fraction: Decimal { self - whole }
}

---

let decimal = Decimal(string: "1234.99999999")!  // 1234.99999999
let fractional = decimal.fraction                // 0.99999999
let whole = decimal.whole                        // 1234
let sum = whole + fractional                     // 1234.99999999

let negativeDecimal = Decimal(string: "-1234.99999999")!  // -1234.99999999
let negativefractional = negativeDecimal.fraction         // -0.99999999
let negativeWhole = negativeDecimal.whole                 // -1234
let negativeSum = negativeWhole + negativefractional      // -1234.99999999

You can use truncatingRemainder and 1 as the divider.

Returns the remainder of this value divided by the given value using truncating division.

Apple doc

Example:

let myDouble1: Double = 12.25
let myDouble2: Double = 12.5
let myDouble3: Double = 12.75

let remainder1 = myDouble1.truncatingRemainder(dividingBy: 1)
let remainder2 = myDouble2.truncatingRemainder(dividingBy: 1)
let remainder3 = myDouble3.truncatingRemainder(dividingBy: 1)

remainder1 -> 0.25
remainder2 -> 0.5
remainder3 -> 0.75