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iOS

Posted by Anthia Tillbury
Anthia Tillbury
3,388 Points

Swift: String Manipulation (Basics)

The following code will complete the second challenge for String Manipulation, but I found a workaround for adding an Integer to a String Literal via the Concatenation method:

let name = "Steven."
let greeting = "Hi there, \(name)"
let finalGreeting = (greeting) + " How are you?"

If changing "name" to an integer the "finalGreeting" Concatenation will include the Integer, believing it to be a String Literal!

let name = 222
let greeting = "Hi there, \(name)"
let finalGreeting = (greeting) + " How are you?"
Martin Wildfeuer
Martin Wildfeuer
Courses Plus Student 11,071 Points

That is expected behavior and actually why String interpolation is such a handy thing: The values wrapped in \() will be converted to a String from any type, it does not have to be a String. From the Apple docs example:

let multiplier = 3
let message = "\(multiplier) times 2.5 is \(Double(multiplier) * 2.5)"
// message is "3 times 2.5 is 7.5"

In the example above, the value of multiplier is inserted into a string literal as (multiplier). This placeholder is replaced with the actual value of multiplier when the string interpolation is evaluated to create an actual string.