Quick way to get current opened filename in a script? I'm looking for an efficient way to get the current .blend file name, in a python script.

I know how to get its path, with bpy.path.abspath('//'), but it returns only the current path, without the file name. I'm sure it's possible, but I can't find out how.
It's in bpy.data.filepath.
This is the absolute path to the currently open blend file, see is_saved, and is_dirty too.