I think you can reproduce the case like this open a new vim :
source the code below : ---------------------------code 1----------------------------- call bufnr( "PV_LISTBUF_8680477943" ,1 ) call setbufvar( 2 , "&modifiable" , 0 ) call setbufvar( 2 , "&swapfile" , 0 ) call setbufvar( 2 , "&buftype" , "nofile" ) call setbufvar( 2 , "&readonly" , 1 ) call setbufvar( 2 , "&bufhidden" , "hide" ) call setbufvar( 2 , "&buflisted" , 0 ) ----------------------------------------------------------------- and now , source these code : -----------------------code 2----------------------------------- echo getbufvar( 2 , "&modifiable") echo getbufvar( 2 , "&swapfile" ) echo getbufvar( 2 , "&buftype" ) echo getbufvar( 2 , "&readonly" ) echo getbufvar( 2 , "&bufhidden" ) echo getbufvar( 2 , "&buflisted" ) ------------------------------------------------------------------ My result is below ( the same as i set by code1 ) : -------------------------result 1--------------------------------- 0 0 nofile 1 hide 0 --------------------------------------------------------------------- The problem occur when i just want to show the buffer do this by :buffer 2 then , source the code2 again At this time , my result is like this : -------------------------result 2 ---------------------------------- 1 0 0 0 ----------------------------------------------------------------------- All my setting to the buffer 2 seems to be discarded. Is there anything wrong with my operation ? PS: version info : vi...@winxp --~--~---------~--~----~------------~-------~--~----~ You received this message from the "vim_dev" maillist. For more information, visit http://www.vim.org/maillist.php -~----------~----~----~----~------~----~------~--~---
