On Wed, Jun 16, 2010 at 8:29 AM, Richard Shaw <hobbes1069@xxxxxxxxx> wrote: > On Wed, Jun 16, 2010 at 7:59 AM, Bruno Wolff III <bruno@xxxxxxxx> wrote: >> On Tue, Jun 15, 2010 at 20:38:56 -0500, >> Richard Shaw <hobbes1069@xxxxxxxxx> wrote: >>> 3 iso_size = math.ceil(file_size/(2*1024))*2*1024 >> >> With loose typing the about might be doing the division as integer division. >> One possible fix would be: >> iso_size = ((file_size + ((2*1024)-1))/(2*1024))*(2*1024) > > I eventually came to the same conclusion last night, I changed that > line to the following: > > iso_size = math.ceil(file_size/(2.0**11))*(2.0**11) > > The "2.0" trips it over to floating point math, but I didn't see any > significant difference in the disc_size calculation. I think I've figured out part of the problem. It looks like if I file is over a certain size, the system reports the size already rounded up by the file system block size. I think my ext4 partition is using a 4k block size. I tried getting the file size of many files, and everything but the smallest files came out even in 4k chunks. I had one small file only at 92 bytes that I get get a file size with significant decimal places. I have no idea how to get around this to get the true file size and then round up the 2k block size of an ISO. Any ideas? Richard -- users mailing list users@xxxxxxxxxxxxxxxxxxxxxxx To unsubscribe or change subscription options: https://admin.fedoraproject.org/mailman/listinfo/users Guidelines: http://fedoraproject.org/wiki/Mailing_list_guidelines
