在异步方法中不能有ref或out参数(如前所述)。

这需要在数据移动中进行一些建模:

public class Data
{
    public int Op {get; set;}
    public int Result {get; set;}
}

public async void Method1()
{
    Data data = await GetDataTaskAsync();
    // use data.Op and data.Result from here on
}

public async Task<Data> GetDataTaskAsync()
{
    var returnValue = new Data();
    // Fill up returnValue
    return returnValue;
}

您获得了更容易重用代码的能力，而且它比变量或元组更具可读性。

你不能有带ref或out参数的异步方法。

Lucian Wischik解释了为什么这是不可能的MSDN线程:http://social.msdn.microsoft.com/Forums/en-US/d2f48a52-e35a-4948-844d-828a1a6deb74/why-async-methods-cannot-have-ref-or-out-parameters

As for why async methods don't support out-by-reference parameters? (or ref parameters?) That's a limitation of the CLR. We chose to implement async methods in a similar way to iterator methods -- i.e. through the compiler transforming the method into a state-machine-object. The CLR has no safe way to store the address of an "out parameter" or "reference parameter" as a field of an object. The only way to have supported out-by-reference parameters would be if the async feature were done by a low-level CLR rewrite instead of a compiler-rewrite. We examined that approach, and it had a lot going for it, but it would ultimately have been so costly that it'd never have happened.

对于这种情况，一个典型的解决方法是让async方法返回一个元组。 你可以像这样重写你的方法:

public async Task Method1()
{
    var tuple = await GetDataTaskAsync();
    int op = tuple.Item1;
    int result = tuple.Item2;
}

public async Task<Tuple<int, int>> GetDataTaskAsync()
{
    //...
    return new Tuple<int, int>(1, 2);
}

在异步方法中不能有ref或out参数(如前所述)。

这需要在数据移动中进行一些建模:

public class Data
{
    public int Op {get; set;}
    public int Result {get; set;}
}

public async void Method1()
{
    Data data = await GetDataTaskAsync();
    // use data.Op and data.Result from here on
}

public async Task<Data> GetDataTaskAsync()
{
    var returnValue = new Data();
    // Fill up returnValue
    return returnValue;
}

您获得了更容易重用代码的能力，而且它比变量或元组更具可读性。

您可以通过使用TPL(任务并行库)而不是直接使用await关键字来做到这一点。

private bool CheckInCategory(int? id, out Category category)
    {
        if (id == null || id == 0)
            category = null;
        else
            category = Task.Run(async () => await _context.Categories.FindAsync(id ?? 0)).Result;

        return category != null;
    }

if(!CheckInCategory(int? id, out var category)) return error

这与Michael Gehling提供的答案非常相似，但我有自己的解决方案，直到我找到了他的解决方案，并注意到我不是第一个想到使用隐式转换的人。

无论如何，当nullable设置为启用时，我也想共享

public readonly struct TryResult<TOut>
{
    #region constructors

    public TryResult(bool success, TOut? value) => (Success, Value) = (success, value);

    #endregion

    #region properties

    public                                            bool  Success { get; init; }
    [MemberNotNullWhen(true, nameof(Success))] public TOut? Value   { get; init; }

    #endregion

    #region methods

    public static implicit operator bool(TryResult<TOut> result) => result.Success;
    public static implicit operator TryResult<TOut>(TOut value) => new (true, value);

    public void Deconstruct(out bool success, out TOut? value) => (success, value) = (Success, Value);

    public TryResult<TOut> Out([NotNullWhen(true)] out TOut? value)
    {
        value = Value;

        return this;
    }

    #endregion
}

然后你可以像这样写一个Try方法:

public static async Task<TryResult<byte[]>> TryGetBytesAsync(string file) =>
    File.Exists(file)
        ? await File.ReadAllBytesAsync(file)
        : default(TryResult<byte[]>);

像这样叫它:

if ((await TryGetBytesAsync(file)).Out(out var bytes))
    Console.WriteLine($"File has {bytes.Length} bytes.");

Alex在可读性上有很好的观点。同样，函数也具有足够的接口来定义返回的类型，并且还可以获得有意义的变量名。

delegate void OpDelegate(int op);
Task<bool> GetDataTaskAsync(OpDelegate callback)
{
    bool canGetData = true;
    if (canGetData) callback(5);
    return Task.FromResult(canGetData);
}

调用者提供一个lambda(或一个命名函数)，智能感知通过从委托复制变量名来提供帮助。

int myOp;
bool result = await GetDataTaskAsync(op => myOp = op);

这种特殊的方法类似于“Try”方法，如果方法结果为真，则设置myOp。否则，你不会关心myOp。