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inout parameter swift – Newbie information – iOSTutorialJunction

By default parameters in swift operate are fixed. It means you cannot change parameters values from inside the operate in swift. Typically we have to get parameters worth to be modified from contained in the operate. 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 operate named swapTwoNumbers, we’re swapping the numbers handed as a parameters to the operate. In the event 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 operate 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 operate utilizing inout parameters

Perform utilizing inout parameters are referred to as like another regular operate 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 are going to 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)")

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

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