PetahChristian PetahChristian - 1 year ago 77
iOS Question

Why must a protocol operator be implemented as a global function?

I've seen the answer to this Swift Equatable Protocol question that mentions how the

method must be declared in the global scope.

If I don't adopt
, I still could declare
to test for equality between two of my types.

// extension Foo: Equatable {}

func ==(lhs: Foo, rhs: Foo) -> Bool {
return ==

struct Foo {
let bar:Int

The fact that its implementation needs to be declared at a global scope, makes it seem incidental to and distinct from a protocol, even if
was adopted.

How is the
protocol anything more than syntactic sugar that merely lets (us and) the compiler safely know that our type implemented the required method of the protocol?

Why does the operator implementation have to be globally declared, even for a protocol? Is this due to some different way that an operator is dispatched?

Answer Source


From the Xcode 8 beta 4 release notes:

Operators can be defined within types or extensions thereof. For example:

struct Foo: Equatable {
    let value: Int
    static func ==(lhs: Foo, rhs: Foo) -> Bool {
        return lhs.value == rhs.value

Such operators must be declared as static (or, within a class, class final), and have the same signature as their global counterparts. As part of this change, operator requirements declared in protocols must also be explicitly declared static:

protocol Equatable {
    static func ==(lhs: Self, rhs: Self) -> Bool


This was discussed on the swift-evolution list recently (2016-01-31 through 2016-02-09 so far). Here's what Chris Lattner said, regarding declaring operators in a struct or class scope:

Yep, this is a generally desirable feature (at least for symmetric operators). This would also be great to get dynamic dispatch of operators within class declarations. I don’t think we have a firm proposal nailing down how name lookup works with this though.

And later (replying to Haravikk):

What are the name lookup issues? Do you mean cases where an operator for Foo == Foo exists in more than one location?

Yes. Name lookup has to have a well defined search order, which defines shadowing and invalid multiple definition rules.

Personally I’d just stick with what we have now, i.e- treat operator implementations within a specific class/struct as being globally defined anyway and throw an error if the same signature is declared more than once.

We need multiple modules to be able to define instances of an operator, we need operators in extensions, and we need retroactive conformance to work, as with any other member.

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