Is Swift Pass By Value or Pass By Reference
Types of Things in Swift
The rule is:
Class instances are reference types (i.e. your reference to a class instance is effectively a pointer)
Functions are reference types
Everything else is a value type; "everything else" simply means instances of structs and instances of enums, because that's all there is in Swift. Arrays and strings are struct instances, for example. You can pass a reference to one of those things (as a function argument) by using
inout
and taking the address, as newacct has pointed out. But the type is itself a value type.
What Reference Types Mean For You
A reference type object is special in practice because:
Mere assignment or passing to function can yield multiple references to the same object
The object itself is mutable even if the reference to it is a constant (
let
, either explicit or implied).A mutation to the object affects that object as seen by all references to it.
Those can be dangers, so keep an eye out. On the other hand, passing a reference type is clearly efficient because only a pointer is copied and passed, which is trivial.
What Value Types Mean For You
Clearly, passing a value type is "safer", and let
means what it says: you can't mutate a struct instance or enum instance through a let
reference. On the other hand, that safety is achieved by making a separate copy of the value, isn't it? Doesn't that make passing a value type potentially expensive?
Well, yes and no. It isn't as bad as you might think. As Nate Cook has said, passing a value type does not necessarily imply copying, because let
(explicit or implied) guarantees immutability so there's no need to copy anything. And even passing into a var
reference doesn't mean that things will be copied, only that they can be if necessary (because there's a mutation). The docs specifically advise you not to get your knickers in a twist.
Everything in Swift is passed by "copy" by default, so when you pass a value-type you get a copy of the value, and when you pass a reference type you get a copy of the reference, with all that that implies. (That is, the copy of the reference still points to the same instance as the original reference.)
I use scare quotes around the "copy" above because Swift does a lot of optimization; wherever possible, it doesn't copy until there's a mutation or the possibility of mutation. Since parameters are immutable by default, this means that most of the time no copy actually happens.
It is always pass-by-value when the parameter is not inout
.
It is always pass-by-reference if the parameter is inout
. However, this is somewhat complicated by the fact you need to explicitly use the &
operator on the argument when passing to an inout
parameter, so it may not fit the traditional definition of pass-by-reference, where you pass the variable directly.
Here is a small code sample for passing by reference. Avoid doing this, unless you have a strong reason to.
func ComputeSomeValues(_ value1: inout String, _ value2: inout Int){
value1 = "my great computation 1";
value2 = 123456;
}
Call it like this
var val1: String = "";
var val2: Int = -1;
ComputeSomeValues(&val1, &val2);
The Apple Swift Developer blog has a post called Value and Reference Types that provides a clear and detailed discussion on this very topic.
To quote:
Types in Swift fall into one of two categories: first, “value types”, where each instance keeps a unique copy of its data, usually defined as a struct, enum, or tuple. The second, “reference types”, where instances share a single copy of the data, and the type is usually defined as a class.
The Swift blog post continues to explain the differences with examples and suggests when you would use one over the other.