CODEWITHSUNDEEP
swiftfunctionrandom
user8537717
user8537717

Reputation:

Swift: how to call the function which calls a random function?

I have four functions which I want to activate randomly. This is the code I have so far, which I get no errors from. However, I get an error "Expression Resolves to an unused function" when I try to call the createSquare() function. 

func createSquare () -> () -> () {
    let squares = [self.squareOne, self.squareTwo, self.squareThree, self.squareFour]

    let randomSquare = Int(arc4random_uniform(UInt32(squares.count)))

    return squares[randomSquare]

}
if gameIsOn == true {
    createSquare()
}

How can I successfully call the function createSquare()?

func squareOne() {
    square1.isHidden = false
}

And squareTwo, squareThree, squareFour are all similar with different numbers.

Upvotes: 1

Views: 120

Answers (2)

yefimovv
yefimovv

Reputation: 7

func createSquare() {
    
    let randomNumber = Int.random(in: 0...3)
    
    if randomNumber == 0 {
        squareOne()
    } else if randomNumber == 1 {
        squareTwo()
    } else if randomNumber == 2 {
        squareThree()
    } else if randomNumber == 3 {
        squareFour()
    }
}

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726849

The problem is that createSquare() returns a closure that you never invoke (that's the "unused function" mentioned in the warning message).

You need to call the function, for example, like this:

let randomFunction = createSquare()
randomFunction()

You can also call it right away with a somewhat less readable syntax:

createSquare()()

Upvotes: 2

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