This is a really specific, technical question (and I think it's fascinating to those of us who don't know the answer) about how the text (code) segment of a program gets loaded into memory. I'm hoping hackers is the right place for this. If not, please forgive and suggest another venue.
Here's my baseline assumption. If I'm wrong here, I'm only going to get wronger as I go:
If I have two different programs that both use a shared library, libfoo.so, the system memory maps the object code they need into the processes' address spaces. There's only one copy of libfoo.so in memory, and the two processes each have handles into it (or maybe just to the pieces of it that they use?).
Step 2 of my question. This gets closer to my real query:
Ok now consider a hard link (not a symlink) from libfoo.so to libbar.so. One inode, two directory entries. Consider my two programs again, one linked against libfoo.so and the other linked against libbar.so. When they run, how many copies of the lib{foo,bar}.so object code are in RAM? My current hypothesis is 1. Isn't it mmapped off the disk? The inode matters, not the file name, right? With me so far? Great.
Now consider jail(8). Let's say I have two jail environments (If you think I mean chroot here, go read jail(8), it's not the same. I'm assuming folks on hackers know jail.). To make my first jail, I make a copy of the FreeBSD stuff that my jail needs. To make the second jail, I create a directory hiearachy, but I *hard link* all the binaries and libraries and stuff to the same inodes that the first jail uses. Is that clear? Let's pick a specific example: 'ln /jail1/usr/sbin/sshd /jail2/usr/sbin/sshd'. Now, sshd uses /usr/lib/libz.so.2. In my example, I have (effectively) done 'ln /jail1/usr/lib/libz.so.2 /jail2/usr/lib/libz.so.2'. These are not symlinks, so this works across jails. Now I launch both jails. Two sshd processes are running, one in each jail.
Now the $64K question: How many instances of, for example, the libz.so.2 object code are in memory? Did my use of jail(8) make any difference? My intuition is that only one copy of the code is in memory for the same reason as in step 2 above. This is the real question I am interested in.
I'm also interested in a broader question. Consider instantiating many jails this way--say 50 or 100 all hard linked to the same base set of files. Can we characterize in some general hand-waving way how much memory (RAM) I would save doing it this way as opposed to the naive way of 50 or 100 copies of the files? I am assuming that if I have 50 copies of the files and I run 50 processes in 50 jails, then I will use more RAM than if I had 50 hard links to the same inode and ran 50 processes in 50 jails from that one inode. The naive copy method will use more RAM, but not 50 times more than the hard linking way.
Thank you to any who respond. I hope I'm not completely out to lunch on this.
Regards, Paco -- Consultant, Cigital, Inc. http://www.cigital.com/
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