"&" in "swap(&someInt, &anotherInt)". What does it represent? What is its function?

Question

I'm new to Swift and I'm trying to learn the concept of 'inout' keyword. I saw this code in "the swift programming language 2.1". My question is, why is there a "&" in "swap(&someInt, &anotherInt)". What does it represent? What is its function?

func swapTwoInts(inout a: Int, inout _ b: Int){
    let temporaryA = a
    a = b
    b = temporaryA
}

var someInt = 3
var anotherInt = 107
swap(&someInt, &anotherInt)

print("someInt is now \(someInt) and anotherInt is now \(anotherInt)")

In the *same section* of the Swift book: *"You place an ampersand (&) directly before a variable’s name when you pass it as an argument to an in-out parameter, to indicate that it can be modified by the function."* — Martin R, Feb 17 '16 at 06:55

Andriko13 · Accepted Answer · 2016-02-17T06:59:39.880

1

Just like passing by reference in C++, the ampersands in the calling bit of the code just tell the compiler that you give permission to function swapTwoInts to change both someInt and anotherInt. If you had not put the ampersands there, the code would not compile.

edited Feb 17 '16 at 06:59

answered Feb 17 '16 at 06:45

Andriko13

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*"If you had not put the ampersands there, the value of both variables would not change after the swap function."* – No, then the code would not compile. – Martin R Feb 17 '16 at 06:50
Indeed, since we need two inout variables. – Andriko13 Feb 17 '16 at 06:59

"&" in "swap(&someInt, &anotherInt)". What does it represent? What is its function?

1 Answers1