CODEWITHSUNDEEP
swiftoption-type
Cilas
Cilas

Reputation: 172

Function returning Optional("String")

I have this function that accepts an Double value, convert it to a currency format and return a String formatted like R$:1.200,30.

func convert_Value(valor: Double) ->String {
    let formatter = NSNumberFormatter()
    formatter.numberStyle = NSNumberFormatterStyle.CurrencyStyle
    formatter.locale = NSLocale(localeIdentifier: "pt_BR")
    return ("\(formatter.stringFromNumber(valor))")
}

This function doesn't have any Optional variable declared, but when i call it using:

x = convert_Value(1200.30)

it returns:

Optional("R$1.200,30")

I can't figure out what i need to do, as its not an optional i can't use exclamation marks to unwrap the optional. I tried to turn the Double and String parameter in function as Optional and then unwrap, but the Optional stills showing.

Upvotes: 2

Views: 1050

Answers (1)

ABakerSmith
ABakerSmith

Reputation: 22939

It doesn't return Optional("R$1.200,30"), it returns "Optional("R$1.200,30")". There's a subtle difference there; notice the ". What's happening is formatter.stringFromNumber(valor) returns String?, which you're putting in a String using "\(...)". Instead, you should return formatter.stringFromNumber(valor)!, force unwrapping here is okay because you know the input is a number.

Upvotes: 6

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