.net - 为什么在 .Net 中使用异步编程模型不会导致 StackOverflow 异常？

Question

例如，我们调用BeginReceive 并具有BeginReceive 在完成时执行的回调方法。如果该回调方法再次在我看来调用 BeginReceive，它将与递归非常相似。这怎么不会导致 stackoverflow 异常。来自 MSDN 的示例代码:

private static void Receive(Socket client) {
    try {
        // Create the state object.
        StateObject state = new StateObject();
        state.workSocket = client;

        // Begin receiving the data from the remote device.
        client.BeginReceive( state.buffer, 0, StateObject.BufferSize, 0,
            new AsyncCallback(ReceiveCallback), state);
    } catch (Exception e) {
        Console.WriteLine(e.ToString());
    }
}

private static void ReceiveCallback( IAsyncResult ar ) {
    try {
        // Retrieve the state object and the client socket 
        // from the asynchronous state object.
        StateObject state = (StateObject) ar.AsyncState;
        Socket client = state.workSocket;

        // Read data from the remote device.
        int bytesRead = client.EndReceive(ar);

        if (bytesRead > 0) {
            // There might be more data, so store the data received so far.
        state.sb.Append(Encoding.ASCII.GetString(state.buffer,0,bytesRead));

            // Get the rest of the data.
            client.BeginReceive(state.buffer,0,StateObject.BufferSize,0,
                new AsyncCallback(ReceiveCallback), state);
        } else {
            // All the data has arrived; put it in response.
            if (state.sb.Length > 1) {
                response = state.sb.ToString();
            }
            // Signal that all bytes have been received.
            receiveDone.Set();
        }
    } catch (Exception e) {
        Console.WriteLine(e.ToString());
    }
}

Answer 1

BeginReceive 注册与重叠 IO 操作关联的回调函数。当数据可用时，操作系统将调用回调，但 BeginReceive 调用会立即返回，因此您对 ReceiveCallback 的调用也会完成。
将实际 IO 视为发生在不属于您的线程中，而是发生在操作系统中。您注册回调的行为只是说“有事发生时继续调用我”，但它不会被添加到堆栈中。这就是它被称为异步的原因。