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>It's a bit wasteful to provision your computers so that all the cold data lives in expensive RAM. But that's a job applications are already doing. They put da

>It's a bit wasteful to provision your computers so that all the cold data lives in expensive RAM. But that's a job applications are already doing. They put data that's being actively worked on in RAM they leave all the rest in storage. Why would you need swap once you can already fit the entire working set in RAM?

eru 1y ago
Sure, some applications are written to manually do a job that your kernel can already do for you. In that case, and if you are only running these applications, the need for swap is much less.
- fluoridation 1y ago
  You mean to tell me most applications you've ever used read the entire file system, loading every file into memory, and rely on the OS to move the unused stuff to swap?
  - eru 1y ago
    No? What makes you think so?
    - fluoridation 1y ago
      Then what do you mean, some applications organize hot and cold data in RAM and storage respectively? Just about every application does it.
      - eru 1y ago
        A silly but realistic example: lots of applications leak a bit of memory here and there. Almost by definition, that leaked memory is never accessed again, so it's very cold. But the applications don't put this on disk by themselves. (If the app's developers knew about which specific bit is leaking, they'd rather fix the leak then write it to disk.)
        fluoridation 1y ago
        That's just recognizing that there's a spectrum of hotness to data. But the question remains: if all the data that the application wants to keep in memory does fit in memory, why do you need swap?
vlovich123 1y ago
Because then you have more active working memory as infrequently used pages are moved to compressed swap and can be used for more page cache or just normal resident memory. Swap ram by itself would be stupid but no one doing this isn’t also turning on compression.
- eru 1y ago
  > Swap ram by itself would be stupid but no one doing this isn’t also turning on compression. I'm not sure what you mean here? Swapping out infrequently accesses pages to disk to make space for more disk cache makes sense with our without compression.
  - vlovich123 1y ago
    Swapping out to RAM without compression is stupid - then you’re just shuffling pages around in memory. Compression is key so that you free up space. Swap to disk is separate.
- fluoridation 1y ago
  >Because then you have more active working memory as infrequently used pages are moved to compressed swap and can be used for more page cache or just normal resident memory. Uhh... A VMM that swaps out to disk an allocated page to make room for more disk cache would be braindead. The process has allocated that memory to use it. The kernel doesn't have enough information to deem disk cache a higher priority. The only thing that should cause it to be swapped out is either another process or the kernel requesting memory.
  - vlovich123 1y ago
    > A VMM that swaps out to disk an allocated page to make room for more disk cache would be braindead Claiming any decision is “brain dead” in something as heuristic heavy and impossible to compute optimally as resident memory pages is quite the statement to make; this is a form of the knapsack problem (NP-complete at least) with the added benefit of time where the items are needed in some specific indeterminate order in the future and there’s a whole bunch of different workloads and workload permutations that alter this. To drive this point home in case you disagree, what’s dumber? Swapping out to disk an allocated page (from the kernel’s perspective) that’s just sitting in the free list of the userspace allocator for that process or a page of some frequently accessed page of data? Now, I agree that VMMs may not do this because it’s difficult to come up with these kinds of scenarios that don’t penalize the general case, more importantly than performance this has to be a mechanism that is explainable to others and understandable for them. But claiming it’s a braindead option to even consider is IMHO a bridge too far.
akvadrako 1y ago
This subthread is about a poster's claim above that every page would be in RAM if you have enough, "hot or not", not just the working set.

eru 1y ago

Sure, some applications are written to manually do a job that your kernel can already do for you. In that case, and if you are only running these applications, the need for swap is much less.

fluoridation 1y ago
You mean to tell me most applications you've ever used read the entire file system, loading every file into memory, and rely on the OS to move the unused stuff to swap?
- eru 1y ago
  No? What makes you think so?
  - fluoridation 1y ago
    Then what do you mean, some applications organize hot and cold data in RAM and storage respectively? Just about every application does it.
    - eru 1y ago
      A silly but realistic example: lots of applications leak a bit of memory here and there. Almost by definition, that leaked memory is never accessed again, so it's very cold. But the applications don't put this on disk by themselves. (If the app's developers knew about which specific bit is leaking, they'd rather fix the leak then write it to disk.)
      - fluoridation 1y ago
        That's just recognizing that there's a spectrum of hotness to data. But the question remains: if all the data that the application wants to keep in memory does fit in memory, why do you need swap?

vlovich123 1y ago

Because then you have more active working memory as infrequently used pages are moved to compressed swap and can be used for more page cache or just normal resident memory. Swap ram by itself would be stupid but no one doing this isn’t also turning on compression.

eru 1y ago
> Swap ram by itself would be stupid but no one doing this isn’t also turning on compression. I'm not sure what you mean here? Swapping out infrequently accesses pages to disk to make space for more disk cache makes sense with our without compression.
- vlovich123 1y ago
  Swapping out to RAM without compression is stupid - then you’re just shuffling pages around in memory. Compression is key so that you free up space. Swap to disk is separate.
fluoridation 1y ago
>Because then you have more active working memory as infrequently used pages are moved to compressed swap and can be used for more page cache or just normal resident memory. Uhh... A VMM that swaps out to disk an allocated page to make room for more disk cache would be braindead. The process has allocated that memory to use it. The kernel doesn't have enough information to deem disk cache a higher priority. The only thing that should cause it to be swapped out is either another process or the kernel requesting memory.
- vlovich123 1y ago
  > A VMM that swaps out to disk an allocated page to make room for more disk cache would be braindead Claiming any decision is “brain dead” in something as heuristic heavy and impossible to compute optimally as resident memory pages is quite the statement to make; this is a form of the knapsack problem (NP-complete at least) with the added benefit of time where the items are needed in some specific indeterminate order in the future and there’s a whole bunch of different workloads and workload permutations that alter this. To drive this point home in case you disagree, what’s dumber? Swapping out to disk an allocated page (from the kernel’s perspective) that’s just sitting in the free list of the userspace allocator for that process or a page of some frequently accessed page of data? Now, I agree that VMMs may not do this because it’s difficult to come up with these kinds of scenarios that don’t penalize the general case, more importantly than performance this has to be a mechanism that is explainable to others and understandable for them. But claiming it’s a braindead option to even consider is IMHO a bridge too far.

akvadrako 1y ago

This subthread is about a poster's claim above that every page would be in RAM if you have enough, "hot or not", not just the working set.