Why doesn't inout pass by reference?

Question

I'm doing something like this:

someFunction(&myClass)

where someFunction sorts an array on myClass.

someFunction(inout someclass:ClassA) {
 someClass.sort({$0.price > $1.price})  
}

If I print myClass after the function call, I notice the array is still unsorted. From what I know, Swift passes values by copy. But when I use inout, shouldn't it change to pass by reference?

That's exactly what it's doing: http://stackoverflow.com/questions/30541244/what-does-a-ampersand-mean-in-the-swift-language — Aidan Gomez, Oct 01 '15 at 05:10
I posted some additional code. @snowman4415's answer makes sense. I tested and that is the case. I'm actually doing a sort() on an array. — 4thSpace, Oct 01 '15 at 13:15
Also note that if you actually were sorting on a class and not an array, you would not need the `inout`, as `inout` is only for assigning a new value to `someclass` and not modifying it in place. Conversely, if you were sorting on a `struct` (likely), you could assign the result to the `inout` param to get the desired effect. — Chris Conover, Oct 06 '15 at 16:09

score 5 · Answer 1 · edited May 23 '17 at 12:16

5

This is because class instances and functions are reference types. Ints, structs, and everything else are value types. When you pass a reference type into a function as a parameter, you are already going to be referencing that instance. When you pass a value type as a parameter, the function gets a copy of that variable (by default), so inout is usually (see edit) only needed if you want to alter a value type from inside of a function.

Altering a class instance without & or inout:

More details

When you create a reference type var t = myClass(), you're really creating a variable t that is a pointer to a myClass instance in memory. By using an ampersand &t in front of a reference type, you are really saying "give me the pointer to the pointer of a myClass instance"

More info on reference vs value types: https://stackoverflow.com/a/27366050/580487

EDIT

As was pointed out in the comments, you can still use inout with reference types if you want to alter a pointer, etc, but I was trying to shed light on the general use case.

Below is an example of sorting an array inside of a function:

edited May 23 '17 at 12:16

Community

1
1

answered Oct 01 '15 at 05:14

tommybananas

5,718
1
28
48

"so inout is needed only if you want to alter a value type from inside of a function." Not true, you need inout with reference types if you want to alter a reference -- i.e. if you want to change it to point to another object. Of course, all reference types are "value types", where the only value is a reference, so technically your statement is correct, but then it would be a trivial statement so that's probably not what you meant. – newacct Oct 01 '15 at 08:34
@snowman4415: I tested and that is the case. I'm actually doing a sort() on an array. I don't see the sorted values when I reference the class after and outside of the function call. – 4thSpace Oct 01 '15 at 13:16
@4thSpace check out my edit. `sort` does not alter the array in place, you need to reassign the return value. – tommybananas Oct 01 '15 at 18:21
2

Please put code examples as text instead of images. Better for copying the code for anyone to try. – Maic López Sáenz Oct 02 '15 at 03:49

score 0 · Answer 2 · answered Oct 01 '15 at 05:24

If you post your code here, it would be more meaningful. BTW, look at below links that might helpful for you,

https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Functions.html#//apple_ref/doc/uid/TP40014097-CH10-ID173

https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Declarations.html#//apple_ref/doc/uid/TP40014097-CH34-ID545

Why doesn't inout pass by reference?

2 Answers2

Linked