Question

我能够使用折叠

来展平列表

flattenWithFold(Iterable list) => list.fold([], (List xs, s) {
s is Iterable ? xs.addAll(flattenWithFold(s)) : xs.add(s);
   return xs;
});

执行时

print(flattenWithFold([1,[3,5,[1,2]],[2,1],6]));

它产生正确的结果[1,3,5,1,2,2,1,6]

但是当我尝试重构使用..add时，它会产生不正确的结果

flattenWithFold1(Iterable list) => list.fold([], (List xs, s) => xs..add(
              s is Iterable ? xs.addAll(flattenWithFold1(s)) : s));

有人可以在执行时解释为什么有空[1,3,5,1,2，null，null，2,1，null，6]？

print(flattenWithFold1([1,[3,5,[1,2]],[2,1],6]));

Answer 1

您的结果中显示null，因为如果您s Iterable xs..add(xs.addAll(flattenWithFold1(s)) addAll。 void是一种null方法，但由于您将其用作表达式，因此会返回xs。因此，您要将展平的元素添加到null，但随后又添加void，这是... gulong clientConnectedConfigureId = g_signal_connect(m_server, "client-connected", (GCallback)ClientConnected, this); ... static void ClientConnected(GstRTSPServer *gstrtspserver, GstRTSPClient *arg1, gpointer user_data) { gulong describeConfigureId = g_signal_connect(arg1, "describe-request", (GCallback)DescribeRequest, user_data); //Do not forget about g_signal_handler_disconnect } void DescribeRequest(GstRTSPClient *gstrtspclient, GstRTSPContext *arg1, gpointer user_data) { gchar *urlString = gst_rtsp_url_get_request_uri(arg1->uri); std::stringstream ssTemp; ssTemp << "DescribeRequest - urlString = " << urlString; InternalLog::Debug(ssTemp.str()); g_free(urlString); }方法的返回值。

折叠展平列表

1 个答案: