void withCheckedThrowingContinuation 无法推断通用参数“T”

Question

withCheckedThrowingContinuation当没有返回类型时，我无法编译。例如：

\n

    public\n    func\n    write(toDevice inDeviceID: Int, atAddress inAddr: Int, value inVal: Float)\n        async\n        throws\n    {\n        try await withCheckedThrowingContinuation\n               // ^ Generic parameter \'T\' could not be inferred\n        { inCont in\n            self.workQ.async\n            {\n                do\n                {\n                    self.deviceID = inDeviceID\n                    try self.write(address: inAddr, value: inVal)\n                    inCont.resume()\n                }\n                \n                catch (let e)\n                {\n                    inCont.resume(throwing: e)\n                }\n            }\n        }\n    }\n

\n

返回的版本工作得很好：

\n

    public\n    func\n    readRegister(address inAddr: Int, fromDevice inDeviceID: Int)\n        async\n        throws\n        -> UInt16\n    {\n        try await withCheckedThrowingContinuation\n        { inCont in\n            self.workQ.async\n            {\n                do\n                {\n                    self.deviceID = inDeviceID\n                    let r = try self.readRegister(address: inAddr)\n                    inCont.resume(returning: r)\n                }\n                \n                catch (let e)\n                {\n                    inCont.resume(throwing: e)\n                }\n            }\n        }\n    }\n

\n

Answer 1

当我写这篇文章时，我发现像这样明确的工作：

    public
    func
    write(toDevice inDeviceID: Int, atAddress inAddr: Int, value inVal: Float)
        async
        throws
    {
        try await withCheckedThrowingContinuation
        { (inCont: CheckedContinuation<Void, Error>) -> Void in
            self.workQ.async
            {
                do
                {
                    self.deviceID = inDeviceID
                    try self.write(address: inAddr, value: inVal)
                    inCont.resume()
                }
                
                catch (let e)
                {
                    inCont.resume(throwing: e)
                }
            }
        }
    }

我向 Apple 提交了一个错误。

Answer 2

如果您return之前使用try await它应该可以工作。我真的不明白为什么，因为在这种情况下不需要退货，但它对我有效。

public
func
write(toDevice inDeviceID: Int, atAddress inAddr: Int, value inVal: Float)
    async
    throws
{
    return try await withCheckedThrowingContinuation
    { inCont in
        self.workQ.async
        {
            do
            {
                self.deviceID = inDeviceID
                try self.write(address: inAddr, value: inVal)
                inCont.resume()
            }
            
            catch (let e)
            {
                inCont.resume(throwing: e)
            }
        }
    }
}

Answer 3

我想我应该添加另一个例子

extension Something {
    public func sync() async throws {
        // The swift compiler demands that we be explicit where there is no return type.
        try await withCheckedThrowingContinuation { (continuation: CheckedContinuation<Void, Error>) -> Void in
            self.sync { result in
                switch result {
                case .success(let entry):
                    continuation.resume(with: .success(entry))
                case .failure(let error):
                    continuation.resume(with: .failure(error))
                }
            }
        }
    }
}

更新：Xcode 现在可以使用 async/await 延续来包装基于闭包的函数。