情况非常简单 - 我写了一个扩展方法,并使其与返回类型 Task<T>
异步 . 但是当我尝试使用await调用它时,编译器会抛出一个错误,表明扩展方法不是我的代码:'t recognized as async at all. Here'
public static async Task<NodeReference<T>> CreateWithLabel<T>(this GraphClient client, T source, String label) where T: class
{
var node = client.Create(source);
var url = string.Format(ConfigurationManager.AppSettings[configKey] + "/node/{0}/labels", node.Id);
var serializedLabel = string.Empty;
using (var tempStream = new MemoryStream())
{
new DataContractJsonSerializer(label.GetType()).WriteObject(tempStream, label);
serializedLabel = Encoding.Default.GetString(tempStream.ToArray());
}
var bytes = Encoding.Default.GetBytes(serializedLabel);
using (var webClient = new WebClient())
{
webClient.Headers.Add("Content-Type", "application/json");
webClient.Headers.Add("Accept", "application/json; charset=UTF-8");
await webClient.UploadDataTaskAsync(url, "POST", bytes);
}
return node;
}
var newNode = await client.CreateWithLabel(new Task(), "Task");
确切的错误信息是这样的:
'await'运算符只能在异步方法中使用 . 考虑使用'async'修饰符标记此方法并将其返回类型更改为'Task'
我做错了什么或是语言/编译器限制?
1 回答
错误消息非常清楚:您调用扩展方法的方法应标记为
async
.现在重读这部分应该更有意义:
“'await'运算符只能用于 within 异步方法 . ”