c# - 将回调方法作为参数传递

Question

我想将回调方法作为参数传递给通用方法，但不知道该怎么做。我试过 Func<IAsyncResult>但它似乎不兼容。

public void webRequest(string apiName, string requestMethod, string requestData, Func<IAsyncResult> callback)
{
    ...
    request.BeginGetRequestStream(new AsyncCallback(GetRequestStreamCallback), request);
}

回调的签名是:

void GetRequestStreamCallback(IAsyncResult asyncResult)

Answer 1

将参数声明为Action<T>而不是 Func<T> .

public void webRequest(string apiName, string requestMethod, string requestData, Action<IAsyncResult> callback)

Func<IAsyncResult>需要一个不带参数并返回 IAsyncResult 的函数实例:

Func<TResult> Delegate

Encapsulates a method that has no parameters and returns a value of the type specified by the TResult parameter.

Action<T>不返回任何东西，只接受参数:

Action<T> Delegate

Encapsulates a method that has a single parameter and does not return a value.