Add generic function implementation to protocol based on another generic function declaration

Paul B 0
Viewed 27

I'd like to create default implementation for convenience protocol function that just calls another protocol function, but can't figure out what's wrong with the code:

protocol RequestManagerProtocol {
    func perform<T: Decodable>(_ request: RequestProtocol) async throws -> (T, Int)
    func perform<T: Decodable>(_ request: RequestProtocol) async throws -> T // Convenience function
}

extension RequestManagerProtocol {
    func perform<T: Decodable>(_ request: RequestProtocol) async throws -> T {
        let (obj, _) = perform<T>(request) // Error: Cannot specialize a non-generic definition
        return obj
    }
}
protocols generics swift
2 Answers

As the error says, you cannot specialize a function. So you cannot do this perform<T>(request), you have to specify the type returned to your constant:

 let (obj, _): (T, Int) = try await perform(request)

Also, since you're not using the returned Int, you can simply do this:

func perform<T: Decodable>(_ request: RequestProtocol) async throws -> T {
    return try await perform(request).0
}
userMtIT0

First of all you have to try await the perform call.

And you have to annotate the (return) type rather than specifying the type in angle brackets

extension RequestManagerProtocol {
    func perform<T: Decodable>(_ request: RequestProtocol) async throws -> T {
        let (obj, _) : (T, Int) = try await perform(request)
        return obj
    }
}
vadian0
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