winapi - How to open a default dialog for window if ShellExecute fails due to no file association in C++?

Question

I can use the windows ShellExecute function to open a file with no problems so long as the file has a correct association.

If no association exists i would like to use the default windows dialog to open the file:

Is this possible? If so how?

Answer 1

The documented way to show that dialog is to use the openas verb.

CoInitializeEx(NULL, COINIT_APARTMENTTHREADED|COINIT_DISABLE_OLE1DDE);
SHELLEXECUTEINFO sei = { sizeof(sei) };
sei.fMask = SEE_MASK_NOASYNC;
sei.nShow = SW_SHOWNORMAL;
sei.lpVerb = "openas";
sei.lpFile = "C:\yourfile.ext";
ShellExecuteEx(&sei);

If you check under HKEY_CLASSES_ROOTUnknownshellopenas you see that this is the same as calling the (undocumented) OpenAs_RunDLL export in shell32.