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'm seeing some strange bugs in my iPhone app that I have narrowed down to my use of NSNumberFormatter.
A stripped down example...
In Xcode playground I have:
let numberFormatter = NSNumberFormatter()
//numberFormatter.numberStyle = .DecimalStyle - does not change behavior
let numberString = "546000.06"
let number: NSNumber = numberFormatter.numberFromString(numberString)!
print("number: \(number)")
let number1: NSNumber = NSDecimalNumber(string: numberString)
print("number1: \(number1)")
This is the output:
number: 546000.0600000001
number1: 546000.06
Note that setting the numberStyle to .DecimalStyle doesn't change anything.
This issue only happens for certain numeric values (for example, 8.03 is another one). I thought NSNumberFormatter was safe for this type of conversion and I haven't seen much noise about this issue on the internet so I want to assume it is something I am doing wrong.
Can anyone explain what I am seeing? Any help is very much appreciated!
It looks like there is an issue with NSNumberFormatter. There are certain values where this rounding error creeps up. In the XCode 7.2.1 playground, it shows up around 8.03.
One way that I've solved this is to round the decimal number. Since the difference is +/- a tiny amount, rounding to 4 fraction places should work. You can use various rounding modes. In this example I used RoundPlain.
var initialValue = NSDecimalNumber(string: "7")
let handler = NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundPlain, scale: 4, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: false)
let numberFormatter = NSNumberFormatter()
for index in 1...300 {
initialValue = initialValue.decimalNumberByAdding(0.01)
let stringValue = "\(initialValue)"
var number = numberFormatter.numberFromString(stringValue)
var decimalNumber: NSDecimalNumber = NSDecimalNumber(decimal: number!.decimalValue)
decimalNumber = decimalNumber.decimalNumberByRoundingAccordingToBehavior(handler)
print("stringValue = \(stringValue), decimalNumber = \(decimalNumber), number = \(number!)")
}
You are seeing a rounding error.
If you would like to display a number to the user with two fraction digits, set maximumFractionDigits:
let numberFormatter = NSNumberFormatter()
numberFormatter.maximumFractionDigits = 2
let firstNumber = NSNumber(float:546000.06)
let firstNumberString = numberFormatter.stringFromNumber(firstNumber)!
let secondNumber = NSNumber(float:5.1337)
let secondNumberString = numberFormatter.stringFromNumber(secondNumber)!
print("firstNumber: \(firstNumberString)")
print("secondNumber: \(secondNumberString)")
The output will be:
"firstNumber: 546000.06\n"
"secondNumber: 5.13\n"
If you try to parse a number from a string, then you are all set.
let thirdNumber = numberFormatter.numberFromString("546000.06")!
print("thirdNumber \(thirdNumber.className): \(thirdNumber)")
The last line prints the description of the object itself (w/ rounding error):
"thirdNumber __NSCFNumber: 546000.0600000001\n"
Update (2016-03-18)
You want to parse the following string:
let currencyString = "$546,000.06"
First, we create a formatter with a locale and .CurrencyStyle. Be aware that the locale depends on the string you try to parse and not on the system you are running on.
let currencyFormatter = NSNumberFormatter.init()
currencyFormatter.locale = NSLocale.init(localeIdentifier: "en_US")
currencyFormatter.numberStyle = .CurrencyStyle
We can now use the formatter to prase the string.
let currencyNumber = currencyFormatter.numberFromString(currencyString)!
You are now free to store this object, for example in CoreData. However, if you wan't to present the value to the user (or print it on the console), you have to use a NSNumberFormatter. If you print the object directly, the description (or debugDescription) of the object is used.
So, let's create another formatter to print the value of your NSNumber object:
let outputFormatter = NSNumberFormatter.init()
outputFormatter.maximumFractionDigits = 10
Using many fraction digits should print any possible rounding errors. But
print("Number: \(outputFormatter.stringFromNumber(currencyNumber)!)")
outputs the desired result:
"Number: 546000.06\n"
What you have seen is the result of the internal representation of an NSNumber object (with description/debugDescription).
I'm seeing some strange bugs in my iPhone app that I have narrowed down to my use of NSNumberFormatter.
A stripped down example...
In Xcode playground I have:
let numberFormatter = NSNumberFormatter()
//numberFormatter.numberStyle = .DecimalStyle - does not change behavior
let numberString = "546000.06"
let number: NSNumber = numberFormatter.numberFromString(numberString)!
print("number: \(number)")
let number1: NSNumber = NSDecimalNumber(string: numberString)
print("number1: \(number1)")
This is the output:
number: 546000.0600000001
number1: 546000.06
Note that setting the numberStyle to .DecimalStyle doesn't change anything.
This issue only happens for certain numeric values (for example, 8.03 is another one). I thought NSNumberFormatter was safe for this type of conversion and I haven't seen much noise about this issue on the internet so I want to assume it is something I am doing wrong.
Can anyone explain what I am seeing? Any help is very much appreciated!
It looks like there is an issue with NSNumberFormatter. There are certain values where this rounding error creeps up. In the XCode 7.2.1 playground, it shows up around 8.03.
One way that I've solved this is to round the decimal number. Since the difference is +/- a tiny amount, rounding to 4 fraction places should work. You can use various rounding modes. In this example I used RoundPlain.
var initialValue = NSDecimalNumber(string: "7")
let handler = NSDecimalNumberHandler(roundingMode: NSRoundingMode.RoundPlain, scale: 4, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: false)
let numberFormatter = NSNumberFormatter()
for index in 1...300 {
initialValue = initialValue.decimalNumberByAdding(0.01)
let stringValue = "\(initialValue)"
var number = numberFormatter.numberFromString(stringValue)
var decimalNumber: NSDecimalNumber = NSDecimalNumber(decimal: number!.decimalValue)
decimalNumber = decimalNumber.decimalNumberByRoundingAccordingToBehavior(handler)
print("stringValue = \(stringValue), decimalNumber = \(decimalNumber), number = \(number!)")
}
You are seeing a rounding error.
If you would like to display a number to the user with two fraction digits, set maximumFractionDigits:
let numberFormatter = NSNumberFormatter()
numberFormatter.maximumFractionDigits = 2
let firstNumber = NSNumber(float:546000.06)
let firstNumberString = numberFormatter.stringFromNumber(firstNumber)!
let secondNumber = NSNumber(float:5.1337)
let secondNumberString = numberFormatter.stringFromNumber(secondNumber)!
print("firstNumber: \(firstNumberString)")
print("secondNumber: \(secondNumberString)")
The output will be:
"firstNumber: 546000.06\n"
"secondNumber: 5.13\n"
If you try to parse a number from a string, then you are all set.
let thirdNumber = numberFormatter.numberFromString("546000.06")!
print("thirdNumber \(thirdNumber.className): \(thirdNumber)")
The last line prints the description of the object itself (w/ rounding error):
"thirdNumber __NSCFNumber: 546000.0600000001\n"
Update (2016-03-18)
You want to parse the following string:
let currencyString = "$546,000.06"
First, we create a formatter with a locale and .CurrencyStyle. Be aware that the locale depends on the string you try to parse and not on the system you are running on.
let currencyFormatter = NSNumberFormatter.init()
currencyFormatter.locale = NSLocale.init(localeIdentifier: "en_US")
currencyFormatter.numberStyle = .CurrencyStyle
We can now use the formatter to prase the string.
let currencyNumber = currencyFormatter.numberFromString(currencyString)!
You are now free to store this object, for example in CoreData. However, if you wan't to present the value to the user (or print it on the console), you have to use a NSNumberFormatter. If you print the object directly, the description (or debugDescription) of the object is used.
So, let's create another formatter to print the value of your NSNumber object:
let outputFormatter = NSNumberFormatter.init()
outputFormatter.maximumFractionDigits = 10
Using many fraction digits should print any possible rounding errors. But
print("Number: \(outputFormatter.stringFromNumber(currencyNumber)!)")
outputs the desired result:
"Number: 546000.06\n"
What you have seen is the result of the internal representation of an NSNumber object (with description/debugDescription).
I am trying to Multiply
self.tipLable.text = String("\((enterBillAmountTextField.text! as NSString).integerValue * (middleTextField.text! as NSString).integerValue * (0.01))")
But getting error Binary operator * cannot be applied to operands of type Int and Double
I am taking values form UITextfields. How to do this multiplication?
extension Double {
// Convert Double to currency
var currency: String {
let formatter = NSNumberFormatter()
formatter.numberStyle = .DecimalStyle
formatter.maximumFractionDigits = 2
formatter.minimumFractionDigits = 2
return formatter.stringFromNumber(self) ?? "0"
}
}
tipLable.text = [enterBillAmountTextField, middleTextField].reduce(0.01) { $0 * (Double($1.text!) ?? 0) }.currency
A slightly shorter and clearer solution. Added a "currency" extension so it can still be done in one line :).
This works
self.tipLable.text = String("\( Double((enterBillAmountTextField.text! as NSString).integerValue) * Double((middleTextField.text! as NSString).integerValue) * 0.01)")
Swift doesn't know how to multiply an Int and a Double. Should the result be an Int or a Double?
Swift won't do implicit type conversion between different operands.
If you want your result to be a Double, both operands should be a Double. Then convert the Double to a String.
While this can all be concisely expressed in one single very long line, perhaps it's more readable and maintainable if you break it out into separate lines:
let subTotal = Double(billAmountTextField.text!) ?? 0
let percent = (Double(middleTextField.text!) ?? 0) * 0.01
let tip = subTotal * percent
self.tipLable.text = String(format: "%.2f", tip) // Handle rounding
The answer you gave is going to bring nightmare to you in some moment.
Try to keep yourself doing things in a way you can guarantee that you are going to be able to test it and that you/or others are going to be able to understand what you are doing there.
/**
Use this function to calculate tip, useful for later testing
- returns: Double Value of the tip you should give
*/
func calculateTip(billAmount billAmount:Double, middleValue:Double) -> Double {
/// Actually calculate Tip if everything is OK
return billAmount * middleValue * 0.01
}
Then in your #IBAction make sure you have correct data before asking
your function for a tip
/// If you have bill data, obtain Double value stored there,
/// if something fails, you should return nil
guard let billAmountText = enterBillAmountTextField.text, billAmount = Double(billAmountText) else {
return
}
/// If you have middle value data, obtain Double value stored there,
/// if something fails, you should return nil
guard let middleText = middleTextField.text, middleValue = Double(middleText) else {
return
}
Then you can call that function
let tip = calculateTip(billAmount: billAmount, middleValue: middleValue).description
//and return in proper format
tipLabel.text = String(format: "%.2f", tip)