在C#中使用通用约束返回Task的异步方法

我已经在a project I’m working on中实现了命令模式.这几乎是当前结构：

public class Response
{
    public bool Success { get; private set; }

    public static Response CreateErrorResponse()
    {
        return new Response { Success = false };
    }
}

public interface ICommand<T> where T : Response
{
    Task<T> ExecuteAsync();
}

public abstract CommandBase : ICommand<T> where T: Response
{
    protected abstract Uri BuildUrl();
    protected abstract Task<T> HandleResponseAsync();

    public async override Task<T> ExecuteAsync()
    {
        var url = BuildUrl();
        var httpClient = new HttpClient();

        var response = await httpClient.GetAsync(url);
        return await HandleResponseAsync(response);
    }
}

我想处理HttpClient可能引发的任何异常,因此我想将CommandBase.ExecuteAsync更改为类似的内容…

public async override Task<T> ExecuteAsync()
{
    var url = BuildUrl();
    var httpClient = new HttpClient();

    try
    {
        var response = await httpClient.GetAsync(url);
        return await HandleResponseAsync(response);
    }
    catch (HttpRequestException hex)
    {
        return Response.CreateErrorResponse(); // doesn't compile
    }
}

我得到的编译错误是“无法将类型Response转换为异步返回类型T”.如in this question所述,我无法使用T.CreateErrorResponse().

我该如何解决？

编辑下降投票者：您是否同意在这样的库中捕获异常,问题仍然存在！

解决方法:

尽管我不确定这是否是最佳解决方案(或在您的特定用例中是否可行),但是您可以执行以下操作：

public class Response
{
    public bool Success { get; private set; }
    public ExceptionDispatchInfo ErrorInfo { get; private set; }
    public bool HasFailed
    {
        get { return !Success; }
    }

    public static T CreateErrorResponse<T>(ExceptionDispatchInfo errorInfo) where T : Response, new()
    {
        var response = new T();
        response.Success = false;
        response.ErrorInfo = errorInfo;
        return response;
    }
}

用法：

catch (HttpRequestException hex)
{
    return Response.CreateErrorResponse<T>(ExceptionDispatchInfo.Capture(hex)); // should compile (I did not check)
}

标签：command-pattern,async-await,generics,c
来源： https://codeday.me/bug/20191122/2062798.html