Dean Kent (dkent@realworldtech.com) on 5/14/07 wrote:
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>anon2 (example@example.net) on 5/14/07 wrote:
>---------------------------
>>
>>No it is NOT. I have no idea where you got this impression. The requirements are (as Linus already explained):
>>
>>Virtual address space >= user address space + kernel address space
>>Kernel address space >= physical memory size + memory-mapped I/O size + kernel virtual mappings
>
>It seems odd to me that virtually *every* address must be mapped, whether actually used or not.
>
>>
>>Assuming you want to allow a user virtual address space
>>on the order of physical memory size, virtual memory size should be at least 2x physical memory.
>
>Even if *none* of your apps uses that much memory? Shouldn't an address space
>only be as large as it needs to be? Shouldn't virtual address pages be allocated only when actually needed?
>
>>
>>The resident set size has NOTHING do do with it. The user (virtual) address space
>>contains all mapped addresses, both resident and non-resident. You need space for all of that.
>
>You need space for what you use, it seems to me. If you are allocating pages for
>unused memory, that seems to be a waste of effort to me.

I'm not sure if part of the confusion comes from the difference between a hard split between Kernel/User processes (2GB/2GB on Windows and 1GB/3GB on Linux by default) and page allocation.

Even Windows (which does not enforce all Kernel address space to map to physical memory) has a fixed split as far as I can tell.

If you place the fixed split at too low a memory range (say 100MB in your example) no process could ever use any more than 100MB.

I'm not sure if I'm beating around the bush or stating the obvious here..