Swift - How to convert String to Double

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

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

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.


Swift 4.2+ String to Double

You should use the new type initializers to convert between String and numeric types (Double, Float, Int). It'll return an Optional type (Double?) which will have the correct value or nil if the String was not a number.

Note: The NSString doubleValue property is not recommended because it returns 0 if the value cannot be converted (i.e.: bad user input).

let lessPrecisePI = Float("3.14")

let morePrecisePI = Double("3.1415926536")
let invalidNumber = Float("alphabet") // nil, not a valid number

Unwrap the values to use them using if/let

if let cost = Double(textField.text!) {
    print("The user entered a value price of \(cost)")
} else {
    print("Not a valid number: \(textField.text!)")
}

You can convert formatted numbers and currency using the NumberFormatter class.

let formatter = NumberFormatter()
formatter.locale = Locale.current // USA: Locale(identifier: "en_US")
formatter.numberStyle = .decimal
let number = formatter.number(from: "9,999.99")

Currency formats

let usLocale = Locale(identifier: "en_US")
let frenchLocale = Locale(identifier: "fr_FR")
let germanLocale = Locale(identifier: "de_DE")
let englishUKLocale = Locale(identifier: "en_GB") // United Kingdom
formatter.numberStyle = .currency

formatter.locale = usLocale
let usCurrency = formatter.number(from: "$9,999.99")

formatter.locale = frenchLocale
let frenchCurrency = formatter.number(from: "9999,99€")
// Note: "9 999,99€" fails with grouping separator
// Note: "9999,99 €" fails with a space before the €

formatter.locale = germanLocale
let germanCurrency = formatter.number(from: "9999,99€")
// Note: "9.999,99€" fails with grouping separator

formatter.locale = englishUKLocale
let englishUKCurrency = formatter.number(from: "£9,999.99")

Read more on my blog post about converting String to Double types (and currency).


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

Tags:

Swift

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