无返回的异步/等待函数 | 雨燕5.5

Question

如何从函数中删除 Bool 返回而不出现错误：

Generic parameter 'T' could not be inferred

这是函数：

private func syncDataStore() async throws -> Bool {
    try await withUnsafeThrowingContinuation { continuation in
        Amplify.DataStore.stop { (result) in
            switch(result) {
            case .success:
                Amplify.DataStore.start { (result) in
                    switch(result) {
                    case .success:
                        print("DataStore started")
                        continuation.resume(returning: true)
                    case .failure(let error):
                        print("Error starting DataStore:\(error)")
                        continuation.resume(throwing: error)
                    }
                }
            case .failure(let error):
                print("Error stopping DataStore:\(error)")
                continuation.resume(throwing: error)
            }
        }
    }
}

这就是我尝试做的，但我收到上面提到的错误：

private func syncDataStore() async throws {
    try await withUnsafeThrowingContinuation { continuation in
        Amplify.DataStore.stop { (result) in
            switch(result) {
            case .success:
                Amplify.DataStore.start { (result) in
                    switch(result) {
                    case .success:
                        print("DataStore started")
                        continuation.resume()
                    case .failure(let error):
                        print("Error starting DataStore:\(error)")
                        continuation.resume(throwing: error)
                    }
                }
            case .failure(let error):
                print("Error stopping DataStore:\(error)")
                continuation.resume(throwing: error)
            }
        }
    }
}

老实说，我不知道为什么它会抱怨，没有回报，而且它与任何模型或任何东西都没有关系......

Answer 1

技巧是将返回值转换为withUnsafeThrowingContinuationto Void，如下所示：

try await withUnsafeThrowingContinuation { continuation in
    someAsyncFunction() { error in
        if let error = error { continuation.resume(throwing: error) }
        else { continuation.resume() }
    }
} as Void

Answer 2

这是效果最好的：

try await withUnsafeThrowingContinuation { (continuation: UnsafeContinuation<Void, Error>) in