How to get the integer part and fractional part of a number in Swift

Question

This is a super basic question, but, I can't seem to find an answer in Swift.

Question:

How do I get the whole integer part and fractional part (to the left and right of the decimal point respectively) of a number in Swift 2 and Swift 3? For example, for the number 1234.56789 —

How do I get the integer part 1234.56789 ?

How do I get the fractional part 1234.56789 ?

Possible duplicate of [Getting the decimal part of a double in Swift](https://stackoverflow.com/questions/31396301/getting-the-decimal-part-of-a-double-in-swift) — Grzegorz Krukowski, Jul 04 '17 at 15:26
Thanks, but this only gives the fractional part, not the integer part. — user4806509, Jul 04 '17 at 16:10
@Mark Dickinson For negative numbers, integer is 1234, fractional part is 56789. (Using `abs()` to give absolute number.) — user4806509, Jul 04 '17 at 17:38

Thomas · Answer 1 · 2019-10-30T09:00:35.317

48

You could do simple floor and truncating:

let value = 1234.56789
let double = floor(value) // 1234.0
let integer = Int(double) // 1234
let decimal = value.truncatingRemainder(dividingBy: 1) // 0.56789

edited Oct 30 '19 at 09:00

answered Jul 04 '17 at 15:33

Thomas

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Great! Don't forget to mark as correct if it helped :) – Thomas Jul 11 '17 at 16:19
floor(value) will give value as 1234.0 and not 1234 – Karthick Ramesh Oct 29 '19 at 22:14
1

@KarthickRamesh Yes you are right, I updated the answer to show the correct integer value. – Thomas Oct 30 '19 at 09:01

score 14 · Answer 2 · answered May 03 '19 at 22:31

14

No need for extensions. Swift already has built in function for this.

let aNumber = modf(3.12345)
aNumber.0 // 3.0
aNumber.1 // 0.12345

answered May 03 '19 at 22:31

Just a coder

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This answer would be way more useful if it referred to the Swift docs for modf. Unfortunately, there pretty much aren't any. The best Swift-specific docs I could find were [here](https://www.alphacodingskills.com/swift/notes/swift-math-modf.php), which claim it's in the Foundation library, but the Apple Xcode Developer docs place it in the Kernel math module. As far as I can tell, this is the same `modf` function found in [standard C](https://en.cppreference.com/w/c/numeric/math/modf). – brotskydotcom Oct 04 '21 at 23:00
Use aNumber.0.description.dropLast(2) if you don't want the point zero at the end. – Brian M Jul 24 '22 at 01:34

Trevor · Answer 3 · 2020-03-05T13:22:29.123

13

Swift 4.xm, 5.x complete solution: I credit @Thomas solution also I'd like to add few things which allow us to use separated parts in 2 string part. Especially when we want to use 2 different UILabel for main and decimal part of the number.

Extension below is also managing number quantity of decimal part. I thought it might be useful.

UPDATE: Now, it is working perfectly with negative numbers as well!

public extension Double{
func integerPart()->String{
    let result = floor(abs(self)).description.dropLast(2).description
    let plusMinus = self < 0 ? "-" : ""
    return  plusMinus + result
}
func fractionalPart(_ withDecimalQty:Int = 2)->String{
    let valDecimal = self.truncatingRemainder(dividingBy: 1)
    let formatted = String(format: "%.\(withDecimalQty)f", valDecimal)
    let dropQuantity = self < 0 ? 3:2
    return formatted.dropFirst(dropQuantity).description
}

edited Mar 05 '20 at 13:22

answered Dec 13 '18 at 01:51

Trevor

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12
16

Thanks dude. This is the most powerful solution – Ali Ihsan URAL Dec 14 '18 at 08:37
To support negative numbers, replace the return in the fraction part with this: formatted.drop(while: {$0 != "."}).description – MetaImi Feb 29 '20 at 19:47
@MetaImi I've tried your proposition; it is confusing with positive numbers in my tests. Now I updated the code above for positive and negative numbers are working perfectly. Thank you for your interest. – Trevor Mar 05 '20 at 13:26
In the ```integerPart()``` you at first use the ```abs``` and then you decide if it is a negative number with ```plusMinus```. Why did you use the ```abs``` in the first place? – Starsky Aug 12 '20 at 14:08
I calculate String formatting there, floor needs (+) value, I did long time ago, probably didn't have right result otherwise try and propose a better line if it can work without ```abs``` – Trevor Aug 22 '20 at 15:32

score 5 · Answer 4 · answered Jul 04 '17 at 15:33

5

Convert your number into String later separate string from .

Try this:-

let number:Float = 123.46789
let numberString = String(number)
let numberComponent = numberString.components(separatedBy :".")
let integerNumber = Int(numberComponent [0])
let fractionalNumber = Int(numberComponent [1])

answered Jul 04 '17 at 15:33

Irshad Ahmad

1,363
11
17

Thanks for this answer. – user4806509 Jul 04 '17 at 16:12

score 4 · Answer 5 · answered Jul 04 '17 at 15:36

4

You could do this ->

let x:Double = 1234.5678
let y:Double = Double(Int(x))
let z:Double = x - Double(Int(x))
print("\(x) \(y) \(z)")

Where x is your original value. y is the integer part and z is the fractional part.

Edit

Thomas answer is the one you want ...

answered Jul 04 '17 at 15:36

Anders Cedronius

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Interesting. Thank you. – user4806509 Jul 04 '17 at 16:12

Eugene · Answer 6 · 2020-02-03T01:30:36.867

modf() works bad with numbers > 1 billion

@Trevor behaves like modf()

@Andres Cedronius example behaves like modf() too (i tried modify it)

@Irshad Ahmed solution is nice

there is some convenience convertions

For some reason you need more control, check out https://developer.apple.com/documentation/foundation/numberformatter

extension Double {

    /// 1.00234 -> 1.0
    var integerPart: Double {
        return Double(Int(self))
    }

    /// 1.0012 --> 0.0012
    var fractionPart: Double {
        let fractionStr = "0.\(String(self).split(separator: ".")[1])"
        return Double(fractionStr)!
    }

    /// 1.0012 --> "0.0012"
    var fractionPartString: String {
        return "0.\(String(self).split(separator: ".")[1])"
    }

    /// 1.0012 --> 12
    var fractionPartInteger: Int {
        let fractionStr = "\(String(self).split(separator: ".")[1])"
        return Int(fractionStr)!
    }

}

print(1000.0.integerPart)  // 1000.0
print(1000.0.fractionPart) // 0.0
print(1000.1.integerPart)  // 1000.0
print(1000.2.fractionPart) // 0.2
print(1_000_000_000.1.integerPart) // 1000000000.0
print(100_000_000.13233.fractionPart) // 0.13233

Navdeep Paliwal · Answer 7 · 2020-01-15T06:14:48.653

2

This will definitely work for you in swift 5 and 4

let number = 3.145443
let integerValue = String(format: "%.0f", number)
let integerValue1 = String(format: "%.1f", number)
let integerValue2 = String(format: "%.2f", number)
print(integerValue)
print(integerValue1)
print(integerValue2)

//Output
3
3.1
3.14

edited Jan 15 '20 at 06:14

answered Dec 21 '19 at 06:16

Navdeep Paliwal

349
3
7

score 0 · Answer 8 · answered Dec 31 '20 at 21:03

var specimen0:Double = 1234.56789

var significant0 = Double.IntegerLiteralType(specimen0) // result is an integer
var fractionals0 = specimen0 - Double(significant0)


var specimen1:Float = -1234.56789

var significant1 = Float.IntegerLiteralType(specimen1) // result is an integer
var fractionals1 = specimen1 - Float(significant1)


var specimen2:CGFloat = -1234.56789

var significant2 = CGFloat.IntegerLiteralType(specimen2) // result is an integer
var fractionals2 = specimen2 - CGFloat(significant2)

These are all built in as of Swift 5.3, I am sure even earlier...

With Double (and Single), due to the way Floating numbers work, you're gonna end up with values for the fractional part, with things like: 0.567890000000034, and for single precision: 0.5678711 — rezwits, Dec 31 '20 at 21:05

How to get the integer part and fractional part of a number in Swift

8 Answers8