Can anyone tell me how to round a double value to x number of decimal places in Swift?
I have:
var totalWorkTimeInHours = (totalWorkTime/60/60)
With totalWorkTime being an NSTimeInterval (double) in second.
totalWorkTimeInHours will give me the hours, but it gives me the amount of time in such a long precise number e.g. 1.543240952039......
How do I round this down to, say, 1.543 when I print totalWorkTimeInHours?
You can use Swift's round function to accomplish this.
To round a Double with 3 digits precision, first multiply it by 1000, round it and divide the rounded result by 1000:
let x = 1.23556789
let y = Double(round(1000 * x) / 1000)
print(y) /// 1.236
Unlike any kind of printf(...) or String(format: ...) solutions, the result of this operation is still of type Double.
EDIT:
Regarding the comments that it sometimes does not work, please read this: What Every Programmer Should Know About Floating-Point Arithmetic
Extension for Swift 2
A more general solution is the following extension, which works with Swift 2 & iOS 9:
extension Double {
/// Rounds the double to decimal places value
func roundToPlaces(places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return round(self * divisor) / divisor
}
}
Extension for Swift 3
In Swift 3 round is replaced by rounded:
extension Double {
/// Rounds the double to decimal places value
func rounded(toPlaces places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
Example which returns Double rounded to 4 decimal places:
let x = Double(0.123456789).roundToPlaces(4) // x becomes 0.1235 under Swift 2
let x = Double(0.123456789).rounded(toPlaces: 4) // Swift 3 version
How do I round this down to, say, 1.543 when I print totalWorkTimeInHours?
To round totalWorkTimeInHours to 3 digits for printing, use the String constructor which takes a format string:
print(String(format: "%.3f", totalWorkTimeInHours))
With Swift 5, according to your needs, you can choose one of the 9 following styles in order to have a rounded result from a Double.
#1. Using FloatingPoint rounded() method
In the simplest case, you may use the Double rounded() method.
let roundedValue1 = (0.6844 * 1000).rounded() / 1000
let roundedValue2 = (0.6849 * 1000).rounded() / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#2. Using FloatingPoint rounded(_:) method
let roundedValue1 = (0.6844 * 1000).rounded(.toNearestOrEven) / 1000
let roundedValue2 = (0.6849 * 1000).rounded(.toNearestOrEven) / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#3. Using Darwin round function
Foundation offers a round function via Darwin.
import Foundation
let roundedValue1 = round(0.6844 * 1000) / 1000
let roundedValue2 = round(0.6849 * 1000) / 1000
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#4. Using a Double extension custom method built with Darwin round and pow functions
If you want to repeat the previous operation many times, refactoring your code can be a good idea.
import Foundation
extension Double {
func roundToDecimal(_ fractionDigits: Int) -> Double {
let multiplier = pow(10, Double(fractionDigits))
return Darwin.round(self * multiplier) / multiplier
}
}
let roundedValue1 = 0.6844.roundToDecimal(3)
let roundedValue2 = 0.6849.roundToDecimal(3)
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#5. Using NSDecimalNumber rounding(accordingToBehavior:) method
If needed, NSDecimalNumber offers a verbose but powerful solution for rounding decimal numbers.
import Foundation
let scale: Int16 = 3
let behavior = NSDecimalNumberHandler(roundingMode: .plain, scale: scale, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: true)
let roundedValue1 = NSDecimalNumber(value: 0.6844).rounding(accordingToBehavior: behavior)
let roundedValue2 = NSDecimalNumber(value: 0.6849).rounding(accordingToBehavior: behavior)
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#6. Using NSDecimalRound(_:_:_:_:) function
import Foundation
let scale = 3
var value1 = Decimal(0.6844)
var value2 = Decimal(0.6849)
var roundedValue1 = Decimal()
var roundedValue2 = Decimal()
NSDecimalRound(&roundedValue1, &value1, scale, NSDecimalNumber.RoundingMode.plain)
NSDecimalRound(&roundedValue2, &value2, scale, NSDecimalNumber.RoundingMode.plain)
print(roundedValue1) // returns 0.684
print(roundedValue2) // returns 0.685
#7. Using NSString init(format:arguments:) initializer
If you want to return a NSString from your rounding operation, using NSString initializer is a simple but efficient solution.
import Foundation
let roundedValue1 = NSString(format: "%.3f", 0.6844)
let roundedValue2 = NSString(format: "%.3f", 0.6849)
print(roundedValue1) // prints 0.684
print(roundedValue2) // prints 0.685
#8. Using String init(format:_:) initializer
Swift’s String type is bridged with Foundation’s NSString class. Therefore, you can use the following code in order to return a String from your rounding operation:
import Foundation
let roundedValue1 = String(format: "%.3f", 0.6844)
let roundedValue2 = String(format: "%.3f", 0.6849)
print(roundedValue1) // prints 0.684
print(roundedValue2) // prints 0.685
#9. Using NumberFormatter
If you expect to get a String? from your rounding operation, NumberFormatter offers a highly customizable solution.
import Foundation
let formatter = NumberFormatter()
formatter.numberStyle = NumberFormatter.Style.decimal
formatter.roundingMode = NumberFormatter.RoundingMode.halfUp
formatter.maximumFractionDigits = 3
let roundedValue1 = formatter.string(from: 0.6844)
let roundedValue2 = formatter.string(from: 0.6849)
print(String(describing: roundedValue1)) // prints Optional("0.684")
print(String(describing: roundedValue2)) // prints Optional("0.685")
In Swift 5.5 and Xcode 13.2:
let pi: Double = 3.14159265358979
String(format:"%.2f", pi)
Example:
PS.: It still the same since Swift 2.0 and Xcode 7.2
This is a fully worked code
Swift 3.0/4.0/5.0 , Xcode 9.0 GM/9.2 and above
let doubleValue : Double = 123.32565254455
self.lblValue.text = String(format:"%.f", doubleValue)
print(self.lblValue.text)
output - 123
let doubleValue : Double = 123.32565254455
self.lblValue_1.text = String(format:"%.1f", doubleValue)
print(self.lblValue_1.text)
output - 123.3
let doubleValue : Double = 123.32565254455
self.lblValue_2.text = String(format:"%.2f", doubleValue)
print(self.lblValue_2.text)
output - 123.33
let doubleValue : Double = 123.32565254455
self.lblValue_3.text = String(format:"%.3f", doubleValue)
print(self.lblValue_3.text)
output - 123.326
Building on Yogi's answer, here's a Swift function that does the job:
func roundToPlaces(value:Double, places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return round(value * divisor) / divisor
}
In Swift 3.0 and Xcode 8.0:
extension Double {
func roundTo(places: Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
Use this extension like so:
let doubleValue = 3.567
let roundedValue = doubleValue.roundTo(places: 2)
print(roundedValue) // prints 3.56
Swift 4, Xcode 10
yourLabel.text = String(format:"%.2f", yourDecimalValue)
The code for specific digits after decimals is:
var a = 1.543240952039
var roundedString = String(format: "%.3f", a)
Here the %.3f tells the swift to make this number rounded to 3 decimal places.and if you want double number, you may use this code:
// String to Double
var roundedString = Double(String(format: "%.3f", b))
Use the built in Foundation Darwin library
SWIFT 3
extension Double {
func round(to places: Int) -> Double {
let divisor = pow(10.0, Double(places))
return Darwin.round(self * divisor) / divisor
}
}
Usage:
let number:Double = 12.987654321
print(number.round(to: 3))
Outputs: 12.988
If you want to round Double values, you might want to use Swift Decimal so you don't introduce any errors that can crop up when trying to math with these rounded values. If you use Decimal, it can accurately represent decimal values of that rounded floating point value.
So you can do:
extension Double {
/// Convert `Double` to `Decimal`, rounding it to `scale` decimal places.
///
/// - Parameters:
/// - scale: How many decimal places to round to. Defaults to `0`.
/// - mode: The preferred rounding mode. Defaults to `.plain`.
/// - Returns: The rounded `Decimal` value.
func roundedDecimal(to scale: Int = 0, mode: NSDecimalNumber.RoundingMode = .plain) -> Decimal {
var decimalValue = Decimal(self)
var result = Decimal()
NSDecimalRound(&result, &decimalValue, scale, mode)
return result
}
}
Then, you can get the rounded Decimal value like so:
let foo = 427.3000000002
let value = foo.roundedDecimal(to: 2) // results in 427.30
And if you want to display it with a specified number of decimal places (as well as localize the string for the user's current locale), you can use a NumberFormatter:
let formatter = NumberFormatter()
formatter.maximumFractionDigits = 2
formatter.minimumFractionDigits = 2
if let string = formatter.string(for: value) {
print(string)
}
A handy way can be the use of extension of type Double
extension Double {
var roundTo2f: Double {return Double(round(100 *self)/100) }
var roundTo3f: Double {return Double(round(1000*self)/1000) }
}
Usage:
let regularPie: Double = 3.14159
var smallerPie: Double = regularPie.roundTo3f // results 3.142
var smallestPie: Double = regularPie.roundTo2f // results 3.14
This is a sort of a long workaround, which may come in handy if your needs are a little more complex. You can use a number formatter in Swift.
let numberFormatter: NSNumberFormatter = {
let nf = NSNumberFormatter()
nf.numberStyle = .DecimalStyle
nf.minimumFractionDigits = 0
nf.maximumFractionDigits = 1
return nf
}()
Suppose your variable you want to print is
var printVar = 3.567
This will make sure it is returned in the desired format:
numberFormatter.StringFromNumber(printVar)
The result here will thus be "3.6" (rounded). While this is not the most economic solution, I give it because the OP mentioned printing (in which case a String is not undesirable), and because this class allows for multiple parameters to be set.
Either:
Using String(format:):
Typecast Double to String with %.3f format specifier and then back to Double
Double(String(format: "%.3f", 10.123546789))!
Or extend Double to handle N-Decimal places:
extension Double {
func rounded(toDecimalPlaces n: Int) -> Double {
return Double(String(format: "%.\(n)f", self))!
}
}
By calculation
multiply with 10^3, round it and then divide by 10^3...
(1000 * 10.123546789).rounded()/1000
Or extend Double to handle N-Decimal places:
extension Double {
func rounded(toDecimalPlaces n: Int) -> Double {
let multiplier = pow(10, Double(n))
return (multiplier * self).rounded()/multiplier
}
}
I would use
print(String(format: "%.3f", totalWorkTimeInHours))
and change .3f to any number of decimal numbers you need
This is more flexible algorithm of rounding to N significant digits
Swift 3 solution
extension Double {
// Rounds the double to 'places' significant digits
func roundTo(places:Int) -> Double {
guard self != 0.0 else {
return 0
}
let divisor = pow(10.0, Double(places) - ceil(log10(fabs(self))))
return (self * divisor).rounded() / divisor
}
}
// Double(0.123456789).roundTo(places: 2) = 0.12
// Double(1.23456789).roundTo(places: 2) = 1.2
// Double(1234.56789).roundTo(places: 2) = 1200
The best way to format a double property is to use the Apple predefined methods.
mutating func round(_ rule: FloatingPointRoundingRule)
FloatingPointRoundingRule is a enum which has following possibilities
Enumeration Cases:
case awayFromZero
Round to the closest allowed value whose magnitude is greater than or equal to that of the source.
case down
Round to the closest allowed value that is less than or equal to the source.
case toNearestOrAwayFromZero
Round to the closest allowed value; if two values are equally close, the one with greater magnitude is chosen.
case toNearestOrEven
Round to the closest allowed value; if two values are equally close, the even one is chosen.
case towardZero
Round to the closest allowed value whose magnitude is less than or equal to that of the source.
case up
Round to the closest allowed value that is greater than or equal to the source.
var aNumber : Double = 5.2
aNumber.rounded(.up) // 6.0
round a double value to x number of decimal
NO. of digits after decimal
var x = 1.5657676754
var y = (x*10000).rounded()/10000
print(y) // 1.5658
var x = 1.5657676754
var y = (x*100).rounded()/100
print(y) // 1.57
var x = 1.5657676754
var y = (x*10).rounded()/10
print(y) // 1.6
For ease to use, I created an extension:
extension Double {
var threeDigits: Double {
return (self * 1000).rounded(.toNearestOrEven) / 1000
}
var twoDigits: Double {
return (self * 100).rounded(.toNearestOrEven) / 100
}
var oneDigit: Double {
return (self * 10).rounded(.toNearestOrEven) / 10
}
}
var myDouble = 0.12345
print(myDouble.threeDigits)
print(myDouble.twoDigits)
print(myDouble.oneDigit)
The print results are:
0.123
0.12
0.1
Thanks for the inspiration of other answers!
Not Swift but I'm sure you get the idea.
pow10np = pow(10,num_places);
val = round(val*pow10np) / pow10np;
Swift 5
using String method
var yourDouble = 3.12345
//to round this to 2 decimal spaces i could turn it into string
let roundingString = String(format: "%.2f", myDouble)
let roundedDouble = Double(roundingString) //and than back to double
// result is 3.12
but it's more accepted to use extension
extension Double {
func round(to decimalPlaces: Int) -> Double {
let precisionNumber = pow(10,Double(decimalPlaces))
var n = self // self is a current value of the Double that you will round
n = n * precisionNumber
n.round()
n = n / precisionNumber
return n
}
}
and then you can use:
yourDouble.round(to:2)
This seems to work in Swift 5.
Quite surprised there isn't a standard function for this already.
//Truncation of Double to n-decimal places with rounding
extension Double {
func truncate(to places: Int) -> Double {
return Double(Int((pow(10, Double(places)) * self).rounded())) / pow(10, Double(places))
}
}
To avoid Float imperfections use Decimal
extension Float {
func rounded(rule: NSDecimalNumber.RoundingMode, scale: Int) -> Float {
var result: Decimal = 0
var decimalSelf = NSNumber(value: self).decimalValue
NSDecimalRound(&result, &decimalSelf, scale, rule)
return (result as NSNumber).floatValue
}
}
ex.
1075.58 rounds to 1075.57 when using Float with scale: 2 and .down
1075.58 rounds to 1075.58 when using Decimal with scale: 2 and .down
var n = 123.111222333
n = Double(Int(n * 10.0)) / 10.0
Result: n = 123.1
Change 10.0 (1 decimal place) to any of 100.0 (2 decimal place), 1000.0 (3 decimal place) and so on, for the number of digits you want after decimal..
The solution worked for me. XCode 13.3.1 & Swift 5
extension Double {
func rounded(decimalPoint: Int) -> Double {
let power = pow(10, Double(decimalPoint))
return (self * power).rounded() / power
}
}
Test:
print(-87.7183123123.rounded(decimalPoint: 3))
print(-87.7188123123.rounded(decimalPoint: 3))
print(-87.7128123123.rounded(decimalPoint: 3))
Result:
-87.718
-87.719
-87.713
I found this wondering if it is possible to correct a user's input. That is if they enter three decimals instead of two for a dollar amount. Say 1.111 instead of 1.11 can you fix it by rounding? The answer for many reasons is no! With money anything over i.e. 0.001 would eventually cause problems in a real checkbook.
Here is a function to check the users input for too many values after the period. But which will allow 1., 1.1 and 1.11.
It is assumed that the value has already been checked for successful conversion from a String to a Double.
//func need to be where transactionAmount.text is in scope
func checkDoublesForOnlyTwoDecimalsOrLess()->Bool{
var theTransactionCharacterMinusThree: Character = "A"
var theTransactionCharacterMinusTwo: Character = "A"
var theTransactionCharacterMinusOne: Character = "A"
var result = false
var periodCharacter:Character = "."
var myCopyString = transactionAmount.text!
if myCopyString.containsString(".") {
if( myCopyString.characters.count >= 3){
theTransactionCharacterMinusThree = myCopyString[myCopyString.endIndex.advancedBy(-3)]
}
if( myCopyString.characters.count >= 2){
theTransactionCharacterMinusTwo = myCopyString[myCopyString.endIndex.advancedBy(-2)]
}
if( myCopyString.characters.count > 1){
theTransactionCharacterMinusOne = myCopyString[myCopyString.endIndex.advancedBy(-1)]
}
if theTransactionCharacterMinusThree == periodCharacter {
result = true
}
if theTransactionCharacterMinusTwo == periodCharacter {
result = true
}
if theTransactionCharacterMinusOne == periodCharacter {
result = true
}
}else {
//if there is no period and it is a valid double it is good
result = true
}
return result
}
You can add this extension :
extension Double {
var clean: String {
return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: "%.0f", self) : String(format: "%.2f", self)
}
}
and call it like this :
let ex: Double = 10.123546789
print(ex.clean) // 10.12
Here's one for SwiftUI if you need a Text element with the number value.
struct RoundedDigitText : View {
let digits : Int
let number : Double
var body : some View {
Text(String(format: "%.\(digits)f", number))
}
}
//find the distance between two points
let coordinateSource = CLLocation(latitude: 30.7717625, longitude:76.5741449 )
let coordinateDestination = CLLocation(latitude: 29.9810859, longitude: 76.5663599)
let distanceInMeters = coordinateSource.distance(from: coordinateDestination)
let valueInKms = distanceInMeters/1000
let preciseValueUptoThreeDigit = Double(round(1000*valueInKms)/1000)
self.lblTotalDistance.text = "Distance is : \(preciseValueUptoThreeDigit) kms"
I am trying to write a code where I have got two time[hh:min] data(String type). Need to just compare but the challenge is my code undergones some validations before returning the final values. so the assertion fails sometimes stating expected value is [17:04] but actual is [17:05]. Is there any way where we can use concept of Threshold that upto few minutes (say 2 mins) the comparison will still be valid?
Step one is do not store a thing as something that it is not. If these are times, they should be stored as times. Strings are for representation to the users; underlying storage is for reality.
So now let's store our times as date components:
let t1 = DateComponents(hour:17, minute:4)
let t2 = DateComponents(hour:17, minute:5)
Now it's easy to find out how far apart they are:
let cal = Calendar(identifier: .gregorian)
if let d1 = cal.date(from: t1),
let d2 = cal.date(from: t2) {
let diff = abs(d1.timeIntervalSince(d2))
// and now decide what to do
}
You first need to seprate your string to an array, and then you can compare.
/* That two arrays are A1 and A2 */
let minute1 = Int(A1[0])*60+Int(A1[1])
let minute2 = Int(A2[0])*60+Int(A2[1])
This may help you. I think that #Sweeper did not understand that it is a time, not a date.
You can convert your string to minutes, subtract one from another and check if the absolute value is less than the threshold:
extension String {
var time24hToMinutes: Int? {
guard count == 5, let hours = Int(prefix(2)), let minutes = Int(suffix(2)), Array(self)[2] == ":" else { return nil }
return hours * 60 + minutes
}
func time24hCompare(to other: String, threshold: Int = 2) -> Bool {
guard let lhs = time24hToMinutes, let rhs = other.time24hToMinutes else { return false }
return abs(lhs-rhs) < threshold
}
}
Testing:
"17:02".time24hCompare(to: "17:04") // false
"17:03".time24hCompare(to: "17:04") // true
"17:04".time24hCompare(to: "17:04") // true
"17:05".time24hCompare(to: "17:04") // true
"17:06".time24hCompare(to: "17:04") // false
It is recommended to round the decimals but i am facing an scenario where i just need to cut down the precision.
Output: 15.96 to 16.0
Desired output: 15.96 to 15.9
Codes:
var value: AnyObject = dict.valueForKey("XXX")!
var stringVal = NSString(format:"%.1f", value.floatValue)
I thought this will be simple but found tricky. Your thoughts on this is highly appreciated.
Use a NSNumberFormatter and configure its rounding mode accordingly:
let formatter = NSNumberFormatter()
formatter.maximumFractionDigits = 1
formatter.roundingMode = .RoundDown
let s = formatter.stringFromNumber(15.96)
// Result: s = "15.9"
If you need to use the rounded number in future math operations, you can use the following function:
func roundToPlaces(_ value: Double, places: Int, rule: FloatingPointRoundingRule = .toNearestOrAwayFromZero) -> Double {
let divisor = pow(10.0, Double(places))
return (value * divisor).rounded(rule) / divisor
}
You can then call it with
var value: AnyObject = dict.valueForKey("XXX")!
var rounded = roundToPlaces(value.doubleValue, places: 1, rule: .down)
var stringVal = "\(rounded)"
What this actually did was the following:
15.96 * 10.0 = 159.6
floor(159.6) = 159.0
159.0 / 10.0 = 15.9
Caveat: This won't help in situations where you're using scientific precision, i.e.
1.49850e0 --> 1.4e0 // (5 places --> 1 place)
1.39e10 --> 1.3e10 // (3 places --> 1 place)
It will treat all numbers as e0
[update 2018-08-09]
Since it seems like my answer is getting some views, I would like to point out that rounding floating-point numbers by division can introduce errors because of how floating-point numbers are stored in memory. as user #mattt has pointed out elsewhere:
floor(1.5679999 * 1000) / 1000 == 1.5669999999999999
(if you want to get your math on, this paper is a great primer on numbers and computers)
If you need that level of precision, use fixed-point numbers instead. Swift provides the Decimal type for this purpose.
The important thing is to understand your problem. If you're working with money or sensor data, you probably want Decimals. If you're working with computer graphics, you can go with Floats.
Try using this:
var test : AnyObject = "15.96"
var rounded_down = floorf(test.floatValue * 10) / 10;
print(rounded_down)
Here's an updated answer in Swift 5 based on #Clafou 's answer. You can use it as an extension to any data type. Example
extension Double {
func cutToDecimalPlace(_ decimalPlaces: Int) -> String{
let formatter = NumberFormatter()
formatter.maximumFractionDigits = decimalPlaces
formatter.roundingMode = .down
return formatter.string(from: NSNumber(value: self)) ?? ""
}
}
And you can call it like this
let priceValueString = "24.124"
let updatedPriceValue = priceValueString.doubleValue.cutToDecimalPlace(1)
Output will be
24.1
Trying to figure out the 'correct' way to round down decimal numbers in Swift and struggling to set up the C calls correctly (or something) as it is returning a weird result. Here's a snippet from Playground:
import Foundation
func roundTo2(result: UnsafePointer<Double>, number: UnsafePointer<Double>) {
var resultCOP = COpaquePointer(result)
var numberCOP = COpaquePointer(number)
NSDecimalRound(resultCOP, numberCOP, 2, .RoundDown)
}
var from: Double = 1.54762
var to: Double = 0.0
roundTo2(&to, &from)
println("From: \(from), to: \(to)")
Output -> From: 1.54762, to: 1.54761981964356
I was hoping for 1.54. Any pointers would be appreciated.
The rounding process should be pretty straightforward without any wrappers. All we should do -- just call the function NSDecimalRound(_:_:_:_:), described there: https://developer.apple.com/documentation/foundation/1412204-nsdecimalround
import Cocoa
/// For example let's take any value with multiple decimals like this:
var amount: NSDecimalNumber = NSDecimalNumber(value: 453.585879834)
/// The mutable pointer reserves only "one cell" in memory for the
let uMPtr = UnsafeMutablePointer<Decimal>.allocate(capacity: 1)
/// Connect the pointer to the value of amount
uMPtr[0] = amount.decimalValue
/// Let's check the connection between variable/pointee and the poiner
Swift.print(uMPtr.pointee) /// result: 453.5858798339999232
/// One more pointer to the pointer
let uPtr = UnsafePointer<Decimal>.init(uMPtr)
/// Standard function call
NSDecimalRound(uMPtr, uPtr, Int(2), NSDecimalNumber.RoundingMode.bankers)
/// Check the result
Swift.print(uMPtr.pointee as NSDecimalNumber) /// result: 453.59
My solution:
var from: Double = 1.54762
var to: Double = 0.0
let decimalSize = 2.0 //you want to round for 2 digits after decimal point, change to your right value
let k = pow(10.0, decimalSize) //k here is 100
let cent = from*k
/*
get floor (integer) value of this double,
equal or less than 'cent'.You will get 154.
For negative value, it will return-155.
If you want to get -154, you have to use ceil(cent) for cent < 0.
*/
let centRound = floor(cent)
to = centRound/k
println("From: \(from), to: \(to)")
As additional info to HoaParis answer, you can make an extensions for Double so you can call it easily again later:
extension Double{
func roundDown(decimals:Int)->Double{
var from: Double = self
var to: Double = 0.0
let decimalSize = 2.0 //you want to round for 2 digits after decimal point, change to your right value
let k = pow(10.0, Double(decimals)) //k here is 100
var cent = from*k
var centRound = floor(cent) //get floor (integer) value of this double.You will get 154.
to = centRound/k
return to
}
}
var from: Double = 1.54762
from.roundDown(2)// 1.54
from.roundDown(3)// 1.547
Here's another approach (if you just want a fix rounding to 2 digits):
extension Double {
mutating func roundTo2Digits() {
self = NSString(format:"%2.2f", self).doubleValue
}
}
var a:Double = 12.3456
a.roundTo2Digits()
// Playground - noun: a place where people can play
import UIKit
// why rounding double (float) numbers is BAD IDEA
let d1 = 0.04499999999999999 // 0.045
let d2 = d1 + 5e-18 // 0.045 (a 'little bit' bigger)
let dd = d2 - d1 // 0.00000000000000000693889390390723
dd == 5e-18 // false
// this should work by mathematical theory
// and it wokrks ...
// BUT!! the Double DOESN'T means Decimal Number
func round(d: Double, decimalNumbers: UInt) -> Double {
let p = pow(10.0, Double(decimalNumbers))
let s = d < 0.0 ? -1.0 : 1.0
let dabs = p * abs(d) + 0.5
return s * floor(dabs) / p
}
// this works as expected
let r1 = round(d1, 3) // 0.045
let r2 = round(d2, 3) // 0.045
r1 == r2 // true
// this works only in our heads, not in my computer
// as expected too ... :-)
let r11 = round(d1, 2) // 0.04
let r21 = round(d2, 2) // 0.05
r11 == r21 // false
// look at the difference, it is just about the decimal numbers required
// are you able predict such a result?