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iOS Swift Functions and Optionals Parameters and Tuples Decomposing a Tuple

Posted by Cody Iddings
Cody Iddings
1,328 Points

I don't understand the relationship: result.language

So, this was written for the Challange at the end of Tuples.

func greeting(person: String) -> (greeting: String, language: String) {
    let language = "English"
    let greeting = "Hello \(person)"

    return (greeting, language)
}

var result = greeting("Tom")

print(result.language)

I don't understand how the variable "result" is related to the language String.

2 Answers

Steven Deutsch
Steven Deutsch
21,046 Points
Steven Deutsch
Steven Deutsch
21,046 Points

Hey Cody Iddings,

func greeting(person: String) -> (greeting: String, language: String) {
    let language = "English"
    let greeting = "Hello \(person)"

    return (greeting, language)
}

var result = greeting("Tom") 

// here you are creating an instance of your greeting method, 
// and assigning the value to the var "result"

print(result.language)

// here you are accessing the value of language in that instance and printing the value

Good luck!

Cody Iddings
Cody Iddings
1,328 Points

I think I understand what you are saying. The instance I'm creating is wrapped up into the variable "result" and so now accessing value "language" through the variable "result"

thanks Steven!

Pedro Laplaza
Pedro Laplaza
1,960 Points
Pedro Laplaza
Pedro Laplaza
1,960 Points

Hi Cody, i had kind of the same issue, it was a little confusing for me to understand where i was pointing when i named each thing so i created the same thing with completely different names to clarify and works, might help you to understand where the values are taken from

func greeting(person: String) -> (greetingString : String,theLanguage : String ){
    let theStringForLanguage = "English"
    let theStringThatSaysHello = "Hello \(person)"

    return (theStringThatSaysHello ,theStringForLanguage)
}

var result = greeting("Tom")
print(result.theLanguage)