CODEWITHSUNDEEP
cachingmemory-managementtlbpage-fault
Bruce
Bruce

Reputation: 35265

Page fault and dirty pages

I have started reading about CPU caches and I have two questions:

  1. Lets say the CPU receives a page fault and transfers control to the kernel handler. The handler decides to evict a frame in memory which is marked dirty. Lets say the CPU caches are write back with valid and modified bits. Now, the memory content of this frame are stale and the cache contains the latest data. How does the kernel force the caches to flush?

  2. The way the page table entry (PTE) gets marked as dirty is as follows: The TLB has a modify bit which is set when the CPU modifies the page's content. This bit is copied back to the PTE on context switch. If we get a page fault, the PTE might be non-dirty but the TLB entry might have the modified bit set (it has not been copied back yet). How is this situation resolved?

Upvotes: 1

Views: 3487

Answers (1)

BoredBlazer
BoredBlazer

Reputation: 364

As for flushing cache, that's just a privileged instruction. The OS calls the instruction and the hardware begins flushing. There's one instruction for invalidating all values and signaling an immediate flush without write back, and there's another instruction that tells the hardware to write data back before flushing. After the instruction call, the hardware (cache controller and I/O) takes over. There are also privileged instructions that tell the hardware to flush the TLB.

I'm not certain about your second question because it's been a while since I've taken an operating systems course, but my understanding is that in the event of a page fault the page will first be brought into the page table. Any page that is removed depends on available space as well as the page replacement algorithm used. Before that page can be brought in, if the page that it is replacing has the modified bit set it must be written out first so an IO is queued up. If it's not modified, then the page is immediately replaced. Same process for the TLB. If the modified bit is set then before that page is replaced you must write it back out so an IO is queued up and you just have to wait.

Upvotes: 1

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