Round Double to closest 10

You can use the round() function (which rounds a floating point numberto the nearest integral value) and apply a "scale factor" of 10:

func roundToTens(x : Double) -> Int {    return 10 * Int(round(x / 10.0))}

Example usage:

print(roundToTens(4.9))  // 0print(roundToTens(15.1)) // 20

In the second example, 15.1 is divided by ten (1.51), rounded (2.0),converted to an integer (2) and multiplied by 10 again (20).

Swift 3:

func roundToTens(_ x : Double) -> Int {    return 10 * Int((x / 10.0).rounded())}

Alternatively:

func roundToTens(_ x : Double) -> Int {    return 10 * lrint(x / 10.0)}

defining the rounding function as

import Foundationfunc round(_ value: Double, toNearest: Double) -> Double {    return round(value / toNearest) * toNearest}

gives you more general and flexible way how to do it

let r0 = round(1.27, toNearest: 0.25)   // 1.25let r1 = round(325, toNearest: 10)      // 330.0let r3 = round(.pi, toNearest: 0.0001)  // 3.1416

You can also extend FloatingPoint protocol and add an option to choose the rounding rule:

extension FloatingPoint {    func rounded(to value: Self, roundingRule: FloatingPointRoundingRule = .toNearestOrAwayFromZero) -> Self {       (self / value).rounded(roundingRule) * value    }}

let value = 325.0value.rounded(to: 10) // 330 (default rounding mode toNearestOrAwayFromZero)value.rounded(to: 10, roundingRule: .down) // 320