[swift] Swift - How to convert String to Double

I'm trying to write a BMI program in swift language. And I got this problem: how to convert a String to a Double?

In Objective-C, I can do like this:

double myDouble = [myString doubleValue];

But how can I achieve this in Swift language?

Swift 4.0

try this

 let str:String = "111.11"
 let tempString = (str as NSString).doubleValue
 print("String:-",tempString)

we can use CDouble value which will be obtained by myString.doubleValue

1.

let strswift = "12"
let double = (strswift as NSString).doubleValue

2.

var strswift= "10.6"
var double : Double = NSString(string: strswift).doubleValue 

May be this help for you.

Try this:

   var myDouble = myString.bridgeToObjectiveC().doubleValue
   println(myDouble)

NOTE

Removed in Beta 5. This no longer works ?

This is building upon the answer by @Ryu

His solution is great as long as you're in a country where dots are used as separators. By default NSNumberFormatter uses the devices locale. Therefore this will fail in all countries where a comma is used as the default separator (including France as @PeterK. mentioned) if the number uses dots as separators (which is normally the case). To set the locale of this NSNumberFormatter to be US and thus use dots as separators replace the line

return NSNumberFormatter().numberFromString(self)?.doubleValue

with

let numberFormatter = NSNumberFormatter()
numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
return numberFormatter.numberFromString(self)?.doubleValue

Therefore the full code becomes

extension String {
    func toDouble() -> Double? {
        let numberFormatter = NSNumberFormatter()
        numberFormatter.locale = NSLocale(localeIdentifier: "en_US_POSIX")
        return numberFormatter.numberFromString(self)?.doubleValue
    }
}

To use this, just call "Your text goes here".toDouble()

This will return an optional Double?

As @Ryu mentioned you can either force unwrap:

println("The value is \(myDouble!)") // prints: The value is 4.2

or use an if let statement:

if let myDouble = myDouble {
    println("The value is \(myDouble)") // prints: The value is 4.2
}

Another option here is converting this to an NSString and using that:

let string = NSString(string: mySwiftString)
string.doubleValue

Extension with optional locale

Swift 2.2

extension String {
    func toDouble(locale: NSLocale? = nil) -> Double? {
        let formatter = NSNumberFormatter()
        if let locale = locale {
            formatter.locale = locale
        }
        return formatter.numberFromString(self)?.doubleValue
    }
}

Swift 3.1

extension String {
    func toDouble(_ locale: Locale) -> Double {
        let formatter = NumberFormatter()
        formatter.numberStyle = .decimal
        formatter.locale = locale
        formatter.usesGroupingSeparator = true
        if let result = formatter.number(from: self)?.doubleValue {
            return result
        } else {
            return 0
        }
    }
}

For a little more Swift feeling, using NSFormatter() avoids casting to NSString, and returns nil when the string does not contain a Double value (e.g. "test" will not return 0.0).

let double = NSNumberFormatter().numberFromString(myString)?.doubleValue

Alternatively, extending Swift's String type:

extension String {
    func toDouble() -> Double? {
        return NumberFormatter().number(from: self)?.doubleValue
    }
}

and use it like toInt():

var myString = "4.2"
var myDouble = myString.toDouble()

This returns an optional Double? which has to be unwrapped.

Either with forced unwrapping:

println("The value is \(myDouble!)") // prints: The value is 4.2

or with an if let statement:

if let myDouble = myDouble {
    println("The value is \(myDouble)") // prints: The value is 4.2
}

Update: For localization, it is very easy to apply locales to the NSFormatter as follows:

let formatter = NSNumberFormatter()
formatter.locale = NSLocale(localeIdentifier: "fr_FR")
let double = formatter.numberFromString("100,25")

Finally, you can use NSNumberFormatterCurrencyStyle on the formatter if you are working with currencies where the string contains the currency symbol.

What also works:

// Init default Double variable
var scanned: Double()

let scanner = NSScanner(string: "String to Scan")
scanner.scanDouble(&scanned)

// scanned has now the scanned value if something was found.

SWIFT 4

extension String {
    func toDouble() -> Double? {
        let numberFormatter = NumberFormatter()
        numberFormatter.locale = Locale(identifier: "en_US_POSIX")
        return numberFormatter.number(from: self)?.doubleValue
    }
}

my problem was comma so i solve it this way:

extension String {
    var doubleValue: Double {
        return Double((self.replacingOccurrences(of: ",", with: ".") as NSString).doubleValue)
    }
}

Here's an extension method that allows you to simply call doubleValue() on a Swift string and get a double back (example output comes first)

println("543.29".doubleValue())
println("543".doubleValue())
println(".29".doubleValue())
println("0.29".doubleValue())

println("-543.29".doubleValue())
println("-543".doubleValue())
println("-.29".doubleValue())
println("-0.29".doubleValue())

//prints
543.29
543.0
0.29
0.29
-543.29
-543.0
-0.29
-0.29

Here's the extension method:

extension String {
    func doubleValue() -> Double
    {
        let minusAscii: UInt8 = 45
        let dotAscii: UInt8 = 46
        let zeroAscii: UInt8 = 48

        var res = 0.0
        let ascii = self.utf8

        var whole = [Double]()
        var current = ascii.startIndex

        let negative = current != ascii.endIndex && ascii[current] == minusAscii
        if (negative)
        {
            current = current.successor()
        }

        while current != ascii.endIndex && ascii[current] != dotAscii
        {
            whole.append(Double(ascii[current] - zeroAscii))
            current = current.successor()
        }

        //whole number
        var factor: Double = 1
        for var i = countElements(whole) - 1; i >= 0; i--
        {
            res += Double(whole[i]) * factor
            factor *= 10
        }

        //mantissa
        if current != ascii.endIndex
        {
            factor = 0.1
            current = current.successor()
            while current != ascii.endIndex
            {
                res += Double(ascii[current] - zeroAscii) * factor
                factor *= 0.1
                current = current.successor()
           }
        }

        if (negative)
        {
            res *= -1;
        }

        return res
    }
}

No error checking, but you can add it if you need it.

On SWIFT 3, you can use:

if let myDouble = NumberFormatter().number(from: yourString)?.doubleValue {
   print("My double: \(myDouble)")
}

Note: - If a string contains any characters other than numerical digits or locale-appropriate group or decimal separators, parsing will fail. - Any leading or trailing space separator characters in a string are ignored. For example, the strings “ 5”, “5 ”, and “5” all produce the number 5.

Taken from the documentation: https://developer.apple.com/reference/foundation/numberformatter/1408845-number

SWIFT 3

To clear, nowadays there is a default method:

public init?(_ text: String)` of `Double` class.

It can be used for all classes.

let c = Double("-1.0")
let f = Double("0x1c.6")
let i = Double("inf")

, etc.

Swift 2 Update There are new failable initializers that allow you to do this in more idiomatic and safe way (as many answers have noted, NSString's double value is not very safe because it returns 0 for non number values. This means that the doubleValue of "foo" and "0" are the same.)

let myDouble = Double(myString)

This returns an optional, so in cases like passing in "foo" where doubleValue would have returned 0, the failable intializer will return nil. You can use a guard, if-let, or map to handle the Optional<Double>

Original Post: You don't need to use the NSString constructor like the accepted answer proposes. You can simply bridge it like this:

(swiftString as NSString).doubleValue

Or you could do:

var myDouble = Double((mySwiftString.text as NSString).doubleValue)

I find more readable to add an extension to String as follow:

extension String {
    var doubleValue: Double {
        return (self as NSString).doubleValue
    }
}

and then you just could write your code:

myDouble = myString.doubleValue

You can use StringEx. It extends String with string-to-number conversions including toDouble().

extension String {
    func toDouble() -> Double?
}

It verifies the string and fails if it can't be converted to double.

Example:

import StringEx

let str = "123.45678"
if let num = str.toDouble() {
    println("Number: \(num)")
} else {
    println("Invalid string")
}

As already pointed out, the best way to achieve this is with direct casting:

(myString as NSString).doubleValue

Building from that, you can make a slick native Swift String extension:

extension String {
    var doubleValue: Double {
        return (self as NSString).doubleValue
    }
}

This allows you to directly use:

myString.doubleValue

Which will perform the casting for you. If Apple does add a doubleValue to the native String you just need to remove the extension and the rest of your code will automatically compile fine!

I haven't seen the answer I was looking for. I just post in here mine in case it can help anyone. This answer is valid only if you don't need a specific format.

Swift 3

extension String {
    var toDouble: Double {
        return Double(self) ?? 0.0
    }
}

Swift 4

extension String {
    func toDouble() -> Double {
        let nsString = self as NSString
        return nsString.doubleValue
    }
}

Using Scanner in some cases is a very convenient way of extracting numbers from a string. And it is almost as powerful as NumberFormatter when it comes to decoding and dealing with different number formats and locales. It can extract numbers and currencies with different decimal and group separators.

import Foundation
// The code below includes manual fix for whitespaces (for French case)
let strings = ["en_US": "My salary is $9,999.99",
               "fr_FR": "Mon salaire est 9 999,99€",
               "de_DE": "Mein Gehalt ist 9999,99€",
               "en_GB": "My salary is £9,999.99" ]
// Just for referce
let allPossibleDecimalSeparators = Set(Locale.availableIdentifiers.compactMap({ Locale(identifier: $0).decimalSeparator}))
print(allPossibleDecimalSeparators)
for str in strings {
    let locale = Locale(identifier: str.key)
    let valStr = str.value.filter{!($0.isWhitespace || $0 == Character(locale.groupingSeparator ?? ""))}
    print("Value String", valStr)

    let sc = Scanner(string: valStr)
    // we could do this more reliably with `filter` as well
    sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
    sc.locale = locale

    print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
    while !(sc.isAtEnd) {
        if let val = sc.scanDouble() {
            print(val)
        }

    }
}

However, there are issues with separators that could be conceived as word delimiters.

// This doesn't work. `Scanner` just ignores grouping separators because scanner tends to seek for multiple values
// It just refuses to ignore spaces or commas for example.
let strings = ["en_US": "$9,999.99", "fr_FR": "9999,99€", "de_DE": "9999,99€", "en_GB": "£9,999.99" ]
for str in strings {
    let locale = Locale(identifier: str.key)
    let sc = Scanner(string: str.value)
    sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted.union(CharacterSet(charactersIn: locale.groupingSeparator ?? ""))
    sc.locale = locale
    print("Locale \(locale.identifier) grouping separator: \(locale.groupingSeparator ?? "") . Decimal separator: \(locale.decimalSeparator ?? "")")
    while !(sc.isAtEnd) {
        if let val = sc.scanDouble() {
            print(val)
        }

    }
}
//     sc.scanDouble(representation: Scanner.NumberRepresentation) could help if there were .currency case

There is no problem to auto detect locale. Note that groupingSeparator in French locale in string "Mon salaire est 9 999,99€" is not a space, though it may render exactly as space (here it doesn't). Thats why the code below works fine without !$0.isWhitespace characters being filtered out.

let stringsArr = ["My salary is $9,999.99",
                  "Mon salaire est 9 999,99€",
                  "Mein Gehalt ist 9.999,99€",
                  "My salary is £9,999.99" ]

let tagger = NSLinguisticTagger(tagSchemes: [.language], options: Int(NSLinguisticTagger.Options.init().rawValue))
for str in stringsArr {
    tagger.string = str
    let locale = Locale(identifier: tagger.dominantLanguage ?? "en")
    let valStr = str.filter{!($0 == Character(locale.groupingSeparator ?? ""))}
    print("Value String", valStr)

    let sc = Scanner(string: valStr)
    // we could do this more reliably with `filter` as well
    sc.charactersToBeSkipped = CharacterSet.decimalDigits.inverted
    sc.locale = locale

    print("Locale \(locale.identifier) grouping separator: |\(locale.groupingSeparator ?? "")| . Decimal separator: \(locale.decimalSeparator ?? "")")
    while !(sc.isAtEnd) {
        if let val = sc.scanDouble() {
            print(val)
        }

    }
}
// Also will fail if groupingSeparator == decimalSeparator (but don't think it's possible)

Use this code in Swift 2.0

let strWithFloat = "78.65"
let floatFromString = Double(strWithFloat)

In Swift 2.0 the best way is to avoid thinking like an Objective-C developer. So you should not "convert a String to a Double" but you should "initialize a Double from a String". Apple doc over here: https://developer.apple.com/library/ios//documentation/Swift/Reference/Swift_Double_Structure/index.html#//apple_ref/swift/structctr/Double/s:FSdcFMSdFSSGSqSd_

It's an optional init so you can use the nil coalescing operator (??) to set a default value. Example:

let myDouble = Double("1.1") ?? 0.0

Please check it on playground!

let sString = "236.86"

var dNumber = NSNumberFormatter().numberFromString(sString)
var nDouble = dNumber!
var eNumber = Double(nDouble) * 3.7

By the way in my Xcode

.toDouble() - doesn't exist

.doubleValue create value 0.0 from not numerical strings...

As of Swift 1.1, you can directly pass String to const char * parameter.

import Foundation

let str = "123.4567"
let num = atof(str) // -> 123.4567

atof("123.4567fubar") // -> 123.4567

If you don't like deprecated atof:

strtod("765.4321", nil) // -> 765.4321

One caveat: the behavior of conversion is different from NSString.doubleValue.

atof and strtod accept 0x prefixed hex string:

atof("0xffp-2") // -> 63.75
atof("12.3456e+2") // -> 1,234.56
atof("nan") // -> (not a number)
atof("inf") // -> (+infinity)

If you prefer .doubleValue behavior, we can still use CFString bridging:

let str = "0xff"
atof(str)                      // -> 255.0
strtod(str, nil)               // -> 255.0
CFStringGetDoubleValue(str)    // -> 0.0
(str as NSString).doubleValue  // -> 0.0

Swift : 4 and 5

There are possibly two ways to do this:

1. String -> Int -> Double:

   let strNumber = "314"
   if let intFromString = Int(strNumber){
       let dobleFromInt = Double(intFromString)
       print(dobleFromInt)
   }
   

2. String -> NSString -> Double

   let strNumber1 = "314"
   let NSstringFromString = NSString(string: strNumber1)
   let doubleFromNSString = NSstringFromString.doubleValue
   print(doubleFromNSString)
   

Use it anyway you like according to you need of the code.

var stringValue = "55"

var convertToDouble = Double((stringValue as NSString).doubleValue)