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inout parameter swift – Newbie information – iOSTutorialJunction

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By default parameters in swift perform are fixed. It means you cannot change parameters values from within the perform in swift. Generally 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 sort. 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. Should 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

Operate utilizing inout parameters are known as like some 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 ought to be a ‘Var’ i.e. variable worth not a ‘let’ i.e. a continuing worth.

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

Should you run above code you’ll be able to see, that each the numbers, firstNumber and secondNumbers are swapped and result’s printed on xcode console.

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