Question

我想将异步操作委托转换为返回指定值的异步函数委托。我想出了一个扩展方法：

public static Func<Task<TResult>> Return<TResult>(this Func<Task> asyncAction, TResult result)
{
    ArgumentValidate.NotNull(asyncAction, nameof(asyncAction));

    return async () =>
    {
        await asyncAction();
        return result;
    };
}

但是，我的扩展方法很麻烦，因为从动作委托同步传递的异常现在从函数委托异步传递。具体地：

Func<Task> asyncAction = () => { throw new InvalidOperationException(); };
var asyncFunc = asyncAction.Return(42);
var task = asyncFunc();   // exception should be thrown here
await task;               // but instead gets thrown here

有没有办法以同步异常继续同步传递的方式创建此包装器？ ContinueWith是可行的吗？

更新：同步抛出异常的异步操作的具体示例：

public static Task WriteAllBytesAsync(string filePath, byte[] bytes)
{
    if (filePath == null)
        throw new ArgumentNullException(filePath, nameof(filePath));
    if (bytes == null)
        throw new ArgumentNullException(filePath, nameof(bytes));

    return WriteAllBytesAsyncInner(filePath, bytes);
}

private static async Task WriteAllBytesAsyncInner(string filePath, byte[] bytes)
{
    using (var fileStream = File.OpenWrite(filePath))
        await fileStream.WriteAsync(bytes, 0, bytes.Length);
}

测试：

Func<Task> asyncAction = () => WriteAllBytesAsync(null, null);
var asyncFunc = asyncAction.Return(42);
var task = asyncFunc();   // ArgumentNullException should be thrown here
await task;               // but instead gets thrown here

Answer 1

好吧，您将无法在初始通话中使用async。这很清楚。但是您可以使用调用该函数的同步委托，然后捕获返回的任务以在异步委托中等待它：

public static Func<Task<TResult>> Return<TResult>(this Func<Task> asyncAction, TResult result)
{
    ArgumentValidate.NotNull(asyncAction, nameof(asyncAction));

    return () =>
    {
        // Call this synchronously
        var task = asyncAction();
        // Now create an async delegate for the rest
        Func<Task<TResult>> intermediate = async () => 
        {
            await task;
            return result;
        };
        return intermediate();
    };
}

或者，将它重构为两个方法，基本上将异步lambda表达式提取为异步方法：

public static Func<Task<TResult>> Return<TResult>(
    this Func<Task> asyncAction, TResult result)
{
    ArgumentValidate.NotNull(asyncAction, nameof(asyncAction));

    return () =>
    {
        var task = asyncAction();
        return AwaitAndReturn(task, result);
    };
}

public static async Func<Task<TResult>> AwaitAndReturn<TResult>(
    this Task asyncAction, TResult result)
{
    await task;
    return result;
}

将异步操作转换为异步函数委托，保留同步异常传递

1 个答案: