SOLVED: Day 7 returning the closure from the function

@young92117

Jan '22

Why the below code is legal swift statement?

func createValidator() -> (String) -> Bool {
    return {
        if $0 == "twostraws" {
            return true
        } else {
            return false
        }
    }
}
let validator = createValidator()
print(validator("twostraws"))

I thought return type is not matched up since closure returns Bool but function returns String.

3

@Obelix

Jan '22

Excellent question!

This is a skill you need to develop and practice!

You need to learn how to read the signature of a function like a sentence. It's a skill, and you need to practice.

func createValidator() -> (String) -> Bool 
     {....snip.....}

Say it OUT LOUD until it makes sense.

createValidator() is a function that returns another function. The function it returns is a function that takes a String as a parameter, and returns a Bool.

Now take a look at the guts of your function.

// what are you returning here?
return {
        // This is a function expressed in closure syntax.
        // $0 means this function requires a parameter. 
        // Because it's compared to "twostraws" 
        // the compiler infers $0 must be a string. This function REQUIRES a string.
        if $0 == "twostraws" {
            return true  // this function returns a Bool
        } else {
            return false // again, returns a Bool. Change this to 42. How does the compiler respond?
        }
    }

Things to note: createValidator() does not take any parameters. createValidator() returns a function

If you hold your mouse over the variable validator and hold down the option key, XCode turns your cursor into a "?". CLICK the variable, and XCode will tell you what type validator is.

Take note! validator is NOT a createValidator() function. Instead, it's the function RETURNED by a createValidator() function.

5

@Obelix

Jan '22

Here's another example. Paste this code into Playgrounds.

Convince yourself that each function has the same signature! Each possible function takes a String and returns a Bool.

// For: young92
// By: Obelix, 23 Jan 2022
// Paste into playgrounds.

func createValidator() -> (String) -> Bool {
    let pickOne = Int.random(in: 1...3)
    // pick a random function to return
    // each one has the same signature!
    if pickOne == 1 {
        return {
            // take a string, return a bool
            print ("Random function #1 \($0)")
            return false
        }
    } else if pickOne == 2 {
        return {
            // take a string, return a bool
            print ("Random function #2 \($0)")
            return true
        }
    } else {
        return {
            // take a string, return a bool
            print ("Random function #3 \($0)")
            return Bool.random() // maybe true? could be false
        }
    }
}
let randomFunction = createValidator()  // what TYPE is randomFunction?
print(randomFunction("feeling lucky?"))

6

@young92117

Jan '22

Thanks for the excellent explanation! It makes much sense now.

3

	@young92117 Jan '22 Why the below code is legal swift statement? `func createValidator() -> (String) -> Bool { return { if $0 == "twostraws" { return true } else { return false } } } let validator = createValidator() print(validator("twostraws"))` I thought return type is not matched up since closure returns Bool but function returns String. 3
	@Obelix Jan '22 Excellent question! This is a skill you need to develop and practice! You need to learn how to read the signature of a function like a sentence. It's a skill, and you need to practice. `func createValidator() -> (String) -> Bool {....snip.....}` Say it OUT LOUD until it makes sense. createValidator() is a function that returns another function. The function it returns is a function that takes a `String` as a parameter, and returns a `Bool`. Now take a look at the guts of your function. `// what are you returning here? return { // This is a function expressed in closure syntax. // $0 means this function requires a parameter. // Because it's compared to "twostraws" // the compiler infers $0 must be a string. This function REQUIRES a string. if $0 == "twostraws" { return true // this function returns a Bool } else { return false // again, returns a Bool. Change this to 42. How does the compiler respond? } }` Things to note: createValidator() does not take any parameters. createValidator() returns a function If you hold your mouse over the variable `validator` and hold down the option key, XCode turns your cursor into a "?". CLICK the variable, and XCode will tell you what type `validator` is. Take note! `validator` is NOT a `createValidator()` function. Instead, it's the function RETURNED by a `createValidator()` function. 5
	@Obelix Jan '22 Here's another example. Paste this code into Playgrounds. Convince yourself that each function has the same signature! Each possible function takes a String and returns a Bool. // For: young92 // By: Obelix, 23 Jan 2022 // Paste into playgrounds. func createValidator() -> (String) -> Bool { let pickOne = Int.random(in: 1...3) // pick a random function to return // each one has the same signature! if pickOne == 1 { return { // take a string, return a bool print ("Random function #1 \($0)") return false } } else if pickOne == 2 { return { // take a string, return a bool print ("Random function #2 \($0)") return true } } else { return { // take a string, return a bool print ("Random function #3 \($0)") return Bool.random() // maybe true? could be false } } } let randomFunction = createValidator() // what TYPE is randomFunction? print(randomFunction("feeling lucky?")) 6
	@young92117 Jan '22 Thanks for the excellent explanation! It makes much sense now. 3

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