Patrick O'Callaghan wrote: > < lots of stuff snipped; if interested, read the thead :-) > > > > I think the solution requires simultaneous knowledge of both > > directories so that two files, one in backup_A and one in backup_B > > that are hardlinked together are counted just once. > > You're right of course. In fact just after posting I thought "shouldn't > that be (du A+B)-(du A)?" but then I saw Roberto's solution so I left it > at that. I don't think (du A+B)-(du A) quite works either, unless by "A" and "B" you mean each of the _individual files_ inside backup_A and backup_B. As in my previous example, if backup_A and backup_B are entirely independent directories with no cross-hardlinks, than the correct result is sizeof[backup_A + backup_B]. For each crosslink file, that filesize must be subtracted, and then the sum of all must be done. So 'du -s -c backup_A backup_B' must, for each file found in backup_B, look to see if it is multiply linked and then check if the other inode is in backup_A. Note that a file in backup_B might be "new" relative to A but linked "forward" to a more recent backup_C, not included on the commandline. So merely looking at number of links is insufficient. `du' has some nice smarts. Dean -- fedora-list mailing list fedora-list@xxxxxxxxxx To unsubscribe: https://www.redhat.com/mailman/listinfo/fedora-list
