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On a 32-bit computer, a virtual memory address is represented as an integer between 0 and 2^32. By virtue of being a 32-bit system, no address can be represented that's lower than 0 or higher than 2^32, and we therefore have a total of 4 GiB (2^32 bytes) of virtual memory to use up. We also know that address spaces are memory protected; they cannot touch, because otherwise one process would be able to “step on the toes” of another. So, if all that I've said is correct, let me now ask this: if we grant that, by Microsoft’s own documentation, 2 GiB of virtual address space are used to operate the system and 2GiB of virtual address space are provided to a single user-mode process, have we not exhausted every possible virtual memory address on a 32-bit system? Would this not mean that we have to resort to disk-swapping just to run 2 processes? Surely, this is too ridiculous to be true, and I just want someone to clarify where my thinking has gone astray...

I have looked at the following questions but none of them seem to give satisfying/consistent/not hand-wavy answers. Or maybe I just don't understand them:

What is the maximum addressable space of virtual memory? - Stack Overflow

Virtual address space in windows - Stack Overflow

What happens when the number of possible virtual addresses are exceeded - Stack Overflow

Thanks! :)

voltamatron
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TL;DR: Virtual memory is per-process and the address space changes when the OS switches execution from one process to another.

Remember that we are talking about virtual memory here and virtual memory is a trick that works because of the cooperation between the OS and the CPU hardware.

The split is not always at 2GB (there is a boot switch for 3GB etc.) but for this discussion lets assume it is always 2GB and that we are on a x86 machine.

When the CPU needs to access a an address in virtual memory it needs to translate the memory page from virtual to physical. The exact mechanics of how this works is too big of a topic to cover here but suffice to say that the translation involves a page directory that has information about present/swapped, modified, COW etc and a way to map the address to physical RAM (and if not present, the CPU will ask the OS to swap it in from the page-file).

The upper 2GB is where the kernel and drivers live and the mapping is always the same in all processes (but it can only be accessed in kernel mode (CPU ring 0)). The lower 2GB however are per-process. Each process have their own set of mappings. When the OS switches executing from one process to another (context switch) the page directory for the CPU the thread is about to run on is changed. This way each process has its own virtual address space.

Anders
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  • Darn it, I don't fully follow...are you saying that the virtual memory addresses are being overwritten after every context switch? Because otherwise I don't see how the OS can guarantee 2 GiB of space to each process when there's like 127 of them concurrently taking up, say, 2.2 GiB of VM out of the 4 our system can maintain. – voltamatron Jun 10 '22 at 03:12
  • Nothing is overwritten. Each process does not have 2gb of memory it can actually use. You need to learn the difference between memory that is actually allocated (backed by RAM or page-file) and the virtual memory address space. The address space is the maximum it could potentially have if it was alone on a system. – Anders Jun 11 '22 at 14:09
  • "The address space is the maximum it could potentially have if it was alone on a system." Thank you! This makes sense! Accepting answer :) – voltamatron Jun 13 '22 at 01:59