I'm trying to separate the decimal and integer parts of a double in swift. I've tried a number of approaches but they all run into the same issue...
let x:Double = 1234.5678
let n1:Double = x % 1.0 // n1 = 0.567800000000034
let n2:Double = x - 1234.0 // same result
let n3:Double = modf(x, &integer) // same result
Is there a way to get 0.5678 instead of 0.567800000000034 without converting to the number to a string?
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
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
Swift 2:
You can use:
modf(x).1
or
x % floor(abs(x))
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
Swift 5.1
let x:Double = 1234.5678
let decimalPart:Double = x.truncatingRemainder(dividingBy: 1) //0.5678
let integerPart:Double = x.rounded(.towardZero) //1234
Both of these methods return Double value.
if you want an integer number as integer part, you can just use
Int(x)
Use Float since it has less precision digits than Double
let x:Double = 1234.5678
let n1:Float = Float(x % 1) // n1 = 0.5678
There’s a function in C’s math library, and many programming languages, Swift included, give you access to it. It’s called modf, and in Swift, it works like this
// modf returns a 2-element tuple,
// with the whole number part in the first element,
// and the fraction part in the second element
let splitPi = modf(3.141592)
splitPi.0 // 3.0
splitPi.1 // 0.141592
You can create an extension like below,
extension Double {
func getWholeNumber() -> Double {
return modf(self).0
}
func getFractionNumber() -> Double {
return modf(self).1
}
}
You can get the Integer part like this:
let d: Double = 1.23456e12
let intparttruncated = trunc(d)
let intpartroundlower = Int(d)
The trunc() function truncates the part after the decimal point and the Int() function rounds to the next lower value. This is the same for positive numbers but a difference for negative numbers. If you subtract the truncated part from d, then you will get the fractional part.
func frac (_ v: Double) -> Double
{
return (v - trunc(v))
}
You can get Mantissa and Exponent of a Double value like this:
let d: Double = 1.23456e78
let exponent = trunc(log(d) / log(10.0))
let mantissa = d / pow(10, trunc(log(d) / log(10.0)))
Your result will be 78 for the exponent and 1.23456 for the Mantissa.
Hope this helps you.
It's impossible to create a solution that will work for all Doubles. And if the other answers ever worked, which I also believe is impossible, they don't anymore.
let _5678 = 1234.5678.description.drop { $0 != "." } .description // ".5678"
Double(_5678) // 0.5678
let _567 = 1234.567.description.drop { $0 != "." } .description // ".567"
Double(_567) // 0.5669999999999999
extension Double {
/// Gets the decimal value from a double.
var decimal: Double {
Double("0." + string.split(separator: ".").last.string) ?? 0.0
}
var string: String {
String(self)
}
}
This appears to solve the Double precision issues.
Usage:
print(34.46979988898988.decimal) // outputs 0.46979988898988
print(34.46.decimal) // outputs 0.46
I want to be able to take a longitude/latitude value of 19.3563443 and turn it into 19.35634 (five decimal places). Is there any easy way to do this?
You can use NumberFormatter:
extension Formatter {
static let number = NumberFormatter()
}
extension FloatingPoint {
func fractionDigitsRounded(to digits: Int, roundingMode: NumberFormatter.RoundingMode = .halfEven) -> String {
Formatter.number.roundingMode = roundingMode
Formatter.number.minimumFractionDigits = digits
Formatter.number.maximumFractionDigits = digits
return Formatter.number.string(for: self) ?? ""
}
}
let value = 19.3563443
let roundedToFive = value.fractionDigitsRounded(to: 5) // "19.35634"
Round, get double value back.
NSString(format: "%.5f", myfloat).doubleValue
This won't truncate, but may be accurate enough for long/lat to 5 decimal places and is quick.
This is a math problem, not a coding problem. Converting between String and Double is not very attractive. In mathematics, truncating digits to the right of the decimal is called reducing scale. To truncate a non-integer, all you need to do is move the decimal to the right however many places you want it truncated (multiply it by 10 to move it one, multiply it by 100 to move it two, etc.), convert it into an integer (which drops everything after the new decimal), and then divide it back by the original multiplier to return the decimal to its original position.
To reduce pi's scale to 3:
move the decimal right: 3.1415927 * 10^3 = 3141.5927
zero the fraction = 3141.0
move the decimal back: 3141 / 10^3 = 3.141
Add it as an extension to Double for convenience:
extension Double {
/// Reduces a double's scale (truncates digits right of the decimal).
func reduceScale(to places: Int) -> Double {
guard !self.isNaN,
!self.isInfinite else {
return 0 // if the double is not a number or infinite, this func will throw an exception error
}
let multiplier = pow(10, Double(places))
let newDecimal = multiplier * self // move the decimal right
let truncated = Double(Int(newDecimal)) // drop the fraction
let originalDecimal = truncated / multiplier // move the decimal back
return originalDecimal
}
}
let dirty = 3.1415927
let clean = dirty.reduceScale(to: 2) // 3.14
I suggest using NSDecimalNumber. I don't know if this is actually better or worse than #Leonardo's answer. But when dealing with numbers I prefer to stick to numbers, and not convert to/from strings. The code below generates a function for a requested number of digits that can be reused as needed.
This truncates by rounding down. To get normal arithmetic rounding, change it to use .RoundPlain instead of .RoundDown.
func makeRounder(#digits: Int16) -> (Double) -> Double {
class RoundHandler : NSObject, NSDecimalNumberBehaviors {
var roundToDigits : Int16
init(roundToDigits: Int16) {
self.roundToDigits = roundToDigits
}
#objc func scale() -> Int16 {
println("Digits: \(digits)")
return roundToDigits
}
#objc func roundingMode() -> NSRoundingMode {
return NSRoundingMode.RoundDown
}
#objc func exceptionDuringOperation(operation: Selector, error: NSCalculationError, leftOperand: NSDecimalNumber, rightOperand: NSDecimalNumber) -> NSDecimalNumber? {
return nil
}
}
let roundHandler = RoundHandler(roundToDigits:digits)
func roundDigits(original:Double) -> Double {
let decimal = NSDecimalNumber(double: original)
let rounded = decimal.decimalNumberByRoundingAccordingToBehavior(roundHandler)
return rounded.doubleValue
}
println("Digits: \(digits)")
return roundDigits
}
let roundTo5 = makeRounder(digits:Int16(5))
roundTo5(19.3563443)
roundTo5(1.23456789)
I found a different answer on a different site and wanted to share it:
let x = 129.88888888777
let y = Double(round(1000*x)/1000)
y returns 129.888
thanks to swift's "round" function.
let x = (long*100000).rounded(.towardZero)/100000
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"
So, I'm trying to develop a simple game written in Swift, but I'm having trouble doing a pretty simple thing. I can't manage to create a random CGPoint... When using arc4random, a compiler error shows up telling me that I can't use Int32 in a CGPoint. So, Is there any way to do this? Any workaround? Thanks!
can also maybe make use of Swift's extensions of base types to create a reusable set of overloaded functions of CGPoint. Maybe something like:
extension CGPoint {
func random()->CGPoint { return CGPoint(x:Int(arc4random()%1000),y:Int(arc4random()%1000))}
func random(range:Int)->CGPoint {
return CGPoint(x:Int(arc4random()%range),y:Int(arc4random()%range))}
func random(rangeX:Int, rangeY:Int)->CGPoint {
return CGPoint(x:Int(arc4random()%rangeX),y:Int(arc4random()%rangeY))}
}
You can then write random CGPoints like this:
var p = CGPoint.random()
//random x and y with a range of 1000
or
var p = CGPoint.random(range:100)
//random x and y with a range of 100
or
var p = CGPoint.random(rangeX:200, rangeY:400)
//random x up to 200 and random y with a range of up to 400
Granted, I'm not in the Xcode IDE at the moment to check syntax / if it compiles correctly but hope that could be of help :-)
...
//////////////////
Swift 1.2 Update
//////////////////
Seems these type-level function calls are not allowed anymore with extensions...at least for CGPoint; probably because CGPoint is actually a struct and not a class based on the current IOS documentation.
Here's a more in-depth version of my extension that allows for Range types.
This is confirmed working as of XCode 6.4 Beta
(Github repository with Playground file found here:
https://github.com/princetrunks/Random-CGPoint-Extension)
//creates random CGPoints in Swift as of XCode Beta 6.4 (6E7)
extension CGPoint {
/*private functions that help alleviate the ambiguity of the modulo bias
and nested typecasting as well as recycle similar functionality
for either Int or Range type parameter inputs */
private func randomInt(num:Int) ->Int{
return Int(arc4random_uniform(UInt32(num)))
}
private func randomIntFromRange(numRange:Range<Int>) ->Int{
return Int(arc4random_uniform(UInt32((numRange.endIndex - numRange.startIndex) + numRange.startIndex)))
}
//private variable for the default range
private var defaultRange : Int{
get{return 1000}
}
//(a) public variable that creates a default random CGPoint
static var randomPoint = CGPoint.zeroPoint.random()
//(b) default random point creation
func random()->CGPoint { return CGPoint(x:randomInt(defaultRange),y:randomInt(defaultRange))}
//(c) using an Int parameter for both the random x and y range
func random(range:Int)->CGPoint {
return CGPoint(x:randomInt(range),y:randomInt(range))
}
//(d) allows for the specification of the x and y random range
func random(#rangeX:Int, rangeY:Int)->CGPoint {
return CGPoint(x:randomInt(rangeX),y:randomInt(rangeY))
}
//(e) allows the same functionality as (c) but with a Range<Int> type parameter
func random(range:Range<Int>)->CGPoint {
return CGPoint(x:randomIntFromRange(range), y:randomIntFromRange(range))
}
//(f) allows the same functionality as (d) but with a Range<Int> type parameter
func random(#rangeX:Range<Int>, rangeY:Range<Int> )->CGPoint {
return CGPoint(x:randomIntFromRange(rangeX), y:randomIntFromRange(rangeY))
}
}
Here's how we can test this extension:
//(a)
let r = CGPoint.randomPoint
//(b)
var anotherRandomPoint = r.random()
//(c)
anotherRandomPoint = r.random(1000)
//(d)
anotherRandomPoint = r.random(0...1000)
//(e)
anotherRandomPoint = r.random(rangeX:90, rangeY: 2000)
//(f)
anotherRandomPoint = r.random(rangeX:0...90, rangeY: 0...2000)
// generates 100 random CGPoints between -1000 and 999
for _ in 0...100 {
anotherRandomPoint.random(-1000...1000)
}
hi what about constructing an Int? Int(arc4random())
e.g.
var p = CGPoint(x:Int(arc4random()%1000),y:Int(arc4random()%1000))
Swift 4,5
// Add some range
let minX = 0
let maxX = 100
let minY = 0
let maxY = 100
let randomX = CGFloat.random(in: minX..<maxX)
let randomY = CGFloat.random(in: minY..<maxY)
let random = CGPoint(x: randomX, y: randomY)
Here is an extension on CGPoint to generate random point based on your x,y closed range.
extension CGPoint {
static func randPoint(xRange: ClosedRange<CGFloat>, yRange: ClosedRange<CGFloat>) -> Self {
let x = CGFloat.random(in: xRange)
let y = CGFloat.random(in: yRange)
return .init(x: x, y: y)
}
}
Does anyone know why the last line outputs -4.0 instead of 4.0?
extension Double {
var absoluteValue: Double {
if self > 0.0 {
return self
}
else {
return -1 * self
}
}
}
var minusTwo = -2.0
minusTwo.absoluteValue // 2.0
let minusThree = -3.0
minusThree.absoluteValue // 3.0
-4.0.absoluteValue // -4.0
It’s just being parsed as -(4.0.absoluteValue). This works:
> (-4.0).absoluteValue
$R2: Double = 4
The reason why is because, you missed the subtle operator precedence of the expression.
In your expression:
-4.0.absoluteValue
There are two operators, - (Unary Minus Operator, prefix) and . (method operator). And it seems the precedence of . has overwhelmed -, and the expression was parsed into:
-(4.0.absoluteValue)
Therefore, you get a result of 4.0's absolute value, then flipped it with a unary minus operator, and got -4.0.
With a more in-depth look into the Swift header file, I noticed that the . (method operator) is not in the list of operators. This means . is a more fundamental operator that the language relies on, and cannot be overloaded like other ordinary operators.
I am assuming that . has a higher precedence than anything else, including + - * / =, thus leading to your confusion.
Btw, here is my solution. I made absoluteValue a method since it makes more sense than making it a property.
extension Double {
func absoluteValue() -> Double {
if self < 0 {
return -self
} else {
return self
}
}
}
let myDouble: Double = -7.65
println("The absolute value of \(myDouble) is \(myDouble.absoluteValue()).")
Tested with Swift 4.0.2:
extension Double {
var absoluteValue: Double {
return abs(self)
}
}
var x = -75.34
print(x.absoluteValue) // Prints 75.34
print((-31.337).absoluteValue) // Prints 31.337
Here is my solution:
extension Double{
// Abs property
var absoluteValue: Double{
return Double.abs(self)
}
// Abs method
func absValFunc() ->Double{
return Double.abs(self)
} }
var doubleValue = -36.00
doubleValue.absoluteValue
doubleValue.absValFunc()