Hello, I am looking at the swapping mechanism in Linux. I have read the relevant chapter 16 in 'Understanding the Linux Kernel' from Bovet&Cesati, and looked at the 2.2.18 kernel source code. I still have the follwing question: Function try_to_swap_out() [p. 481 in 'Understanding the Linux Kernel']: If the page in question already belongs to the swap cache, the function performs no data transfer to the swap space on the disk (but only marks the page as swapped out). The corresponding comment in the try_to_swap_out() functions states 'Is the page already in the swap cache? If so, ..... - it is already up-to-date on disk. Understanding the Linux Kernel states on p. 482 'If the page belongs to the swap cache .... no memory transfer is performed'. Now my question is, couldn't the page have been modified since it was added to the swap cache (and written to disk), and thus differ from the data in the swap space? In this case shouldn't the page be written to disk (again)? Such a modification may result from a store operation of another process that shares the page with the process from which it was added to the swap cache, or by an I/O operation from some external device - in both cases the data stored in the corresponding page slot in the swap area differs from the data stored in the page frame. If later on the page frame is released by shrink_mmap() from the swap cache, and subsequently needs to be restored from the data in the swap area, the page frame's latest content is lost. Thanks in advance for any help with best regards Martin Maletinsky P.S. Please put me on CC: in your reply, since I am not in the mailing list. -- Supercomputing System AG email: maletinsky@scs.ch Martin Maletinsky phone: +41 (0)1 445 16 05 Technoparkstrasse 1 fax: +41 (0)1 445 16 10 CH-8005 Zurich -- Kernelnewbies: Help each other learn about the Linux kernel. Archive: http://mail.nl.linux.org/kernelnewbies/ FAQ: http://kernelnewbies.org/faq/
