如何同步一个弹出窗口始终在另一个弹出窗口上方？

Question

How to sync one popup window is always above another popup window ? Both windows are in the same process but in different threads. Similar method the system uses for owner and owned windows, but such windows must belong same thread. In my case, there are two different UI threads in one process and windows are created in those different threads.

Answer 1

使用SetWindowPos ：

SetWindowPos(window_1, window_2, 0, 0, 0, 0, SWP_NOMOVE|SWP_NOSIZE);

Answer 2

Similar method the system uses for owner and owned windows, but such windows must belong same thread.

This statement is incorrect. You can indeed make the owner of a window be a window from a different process. And that is the correct way to do what you desire.