Hi, > The simple explanation is that the value of umask will be subtracted from > the protection mask of whatever you create. If you are creating an > executable file and your umask is 022 then the resulting protection on the > file will be 777-022==755 > Subtracting umask from default permissions will work in many cases, but not always. If you have a bit set in umask which is reset in corresponding position in default permissions, you'll get wrong results subtracting umask value from default permission value. (for example, if you have 6 in default permission and 3 in umask, the resulting permission will be 4, not 3). To get the right results in all cases, you must calculate default permission value AND (NOT umask value). []'s Marcelo
