Swift - Syntax to create a type that implements a protocol using a certain associated type

Originator:jan.sabbe
Number:rdar://17412293 Date Originated:2014/06/22
Status:Open Resolved:
Product:Developer Tools Product Version:
Classification:Feature (New) Reproducible:Always
 
On Swift programming language:

My use case: see also https://devforums.apple.com/message/991691

Given a protocol 'Specification'
protocol Specification {
    typealias MatchType
    func matches(matchObject: MatchType) -> Bool
}

and an implementation
struct AndSpecification<L : Specification, R: Specification where L.MatchType == R.MatchType>  : Specification {
    var left : L
    var right : R
    func matches(matchObject: L.MatchType) -> Bool {...}
}

I would like to be able to define a function

@infix func && <L: Specification, R: Specification where L.MatchType == R.MatchType> (left: L, right: R) -> protocol<S: Specification where S.MatchType == L.MatchType>  {
    return AndSpecification(left: left, right: right)
}

I could specify an AndSpecification as a result type, but I would prefer to hide this 'AndSpecification' type, and only expose a more abstract type to my callers. The Specification protocol is too generic here, caller should know that the MatchType is same as the MatchType of the arguments

Comments


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