> The one major point of difference is that to make an address allocation in the page table doesn't require a physical allocation
No it does require a physical allocation. The page table entry for the new virtual memory needs to be a physical allocation!
People say 'you can have 256 TB of virtual memory'! Yes you can, but you will need 256 GB of physical memory to hold the page table for that, won't you, assuming 4 KB pages, even if none of that 256 TB is committed to physical memory.
You say 'that's not how the OS's page management tables work' - yes it is! Look up Intel 64 and IA-32 Architectures Software Developer’s Manual, Volume 3, Chapter 4, formats of page-directory entry and page-table entry. It's committed memory.
You don't have to allocate a page on the table for an area of memory that is not physically allocated at all. Yes, accessing it will trap to the operating system, but that's the entire point.
No it does require a physical allocation. The page table entry for the new virtual memory needs to be a physical allocation!
People say 'you can have 256 TB of virtual memory'! Yes you can, but you will need 256 GB of physical memory to hold the page table for that, won't you, assuming 4 KB pages, even if none of that 256 TB is committed to physical memory.
You say 'that's not how the OS's page management tables work' - yes it is! Look up Intel 64 and IA-32 Architectures Software Developer’s Manual, Volume 3, Chapter 4, formats of page-directory entry and page-table entry. It's committed memory.