inout parameters in Swift

Inout parameters are the parameters which hold the reference to the original variable. Hence, any change in those params directly affects the original variable.

*How to accept inout param - * We need to add an additional keyword 'inout' just before the data type of parameter as below.

Below example is a function that swaps the values of the original variables.

func swapValues(x:inout Int,y:inout Int){ let z = x x = y y = z } 

*How to pass inout values - * While passing variable to inout type , we need to add the ampersand (&) with variable name. Check below -

override func viewDidLoad() { super.viewDidLoad() // Do any additional setup after loading the view. print("BeforeSwap-> a->\(a), b->\(b)") swapValues(x: &a, y: &b) print("AfterSwap-> a->\(a), b->\(b)") } 

When you run this code you will see below output -