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inout parameter swift – Newbie information – iOSTutorialJunction

By default parameters in swift perform are fixed. It means you cannot change parameters values from inside the perform in swift. Typically we have to get parameters worth to be modified from contained in the perform. So right here, inout parameters involves rescue as inout parameter in swift act like a reference kind. Take a look at under code block

func swapTwoNumbers(_ numA:Int, _ numbB:Int) {
        let temporaryValue = numA
        numA = numbB
        numbB = temporaryValue

In above perform named swapTwoNumbers, we’re swapping the numbers handed as a parameters to the perform. In case you run the above code, compiler will throw an error “Cannont assign to worth: ‘numA’ is a ‘let’fixed”. As a result of parameters handed in a swift perform are fixed by default.

Cannont assign to value:  'numA' is a 'let'constant" - inout parameter swift

Utilizing inout parameter

So right here, to take away the above error we are able to use inout key phrase and make these parameters as inout parameter in swift. Allow us to see how we are able to do it

 func swapTowNumbers(_ numA: inout Int, _ numbB:inout Int) {
        let temporaryValue = numA
        numA = numbB
        numbB = temporaryValue

Name a perform utilizing inout parameters

Perform utilizing inout parameters are referred to as like every other regular perform name. The one distinction is, we have to add ‘&’ earlier than the inout parameter title or worth. ‘&’ signifies that we this parameter is an inout parameter. Secondly, the values we’ll go as parameter needs to be a ‘Var’ i.e. variable worth not a ‘let’ i.e. a relentless worth.

    	var firstNumber = 10
        var secondNumber = 30
        swapTowNumbers(&firstNumber, &secondNumber)
        print("Firstnumber == (firstNumber), and secondnumber == (secondNumber)")

In case you run above code you possibly can see, that each the numbers, firstNumber and secondNumbers are swapped and result’s printed on xcode console.

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