3 ms·

As you observed before, real mode is also privileged. Both before and immediately after enabling protected mode, the code segment you are running from is alread

As you observed before, real mode is also privileged. Both before and immediately after enabling protected mode, the code segment you are running from is already "ring 0" (even if it wasn't loaded from a GDT descriptor), and so it is allowed to transfer control to the "proper" kernel code segment.

bogomipz 5y ago
Thank you. For clarification is that same code segment from real mode sort of automatically "remapped" then to a descriptor in the GDT after CPU transitions into protected mode? Is that correct mental model?
- rep_lodsb 5y ago
  No, how would that even work? The OS (or loader) is responsible for setting up descriptors and managing memory. In 64-bit and most 32-bit kernels, segmentation is basically not used anymore, and the descriptors are just flat address spaces (all start at 0, no limit) with different privilege levels. Example of 32 bit flat GDT: 0000 (null descriptor; reserved by hardware and can't be used) 0008 kernel code (ring 0) 0010 kernel data (ring 0) 0018 user code (ring 3) 0020 user data (ring 3) 0028 TSS (hardware defined structure) Isolation between processes would then be done through the page mapping mechanism. This is kind of hard to wrap your head around at first, but makes it much easier to allocate memory. One big problem with using segments instead of fixed-size pages is that when you need to free them in some random order, memory becomes fragmented. But you can also have segments in addition to paging, which is helpful for transitioning from/to 16-bit mode. Hope that helps at least a little bit!
  - bogomipz 5y ago
    Sorry "remapping" was a poor choice of words on my part. What I meant was the code segment register gets reloaded to some offset in that GDT that has those protection bits set to 0. According to your flat address space map above that would be at offsets 0008 and 0010. That makes a lot of sense, thanks.

bogomipz 5y ago

Thank you. For clarification is that same code segment from real mode sort of automatically "remapped" then to a descriptor in the GDT after CPU transitions into protected mode? Is that correct mental model?

rep_lodsb 5y ago
No, how would that even work? The OS (or loader) is responsible for setting up descriptors and managing memory. In 64-bit and most 32-bit kernels, segmentation is basically not used anymore, and the descriptors are just flat address spaces (all start at 0, no limit) with different privilege levels. Example of 32 bit flat GDT: 0000 (null descriptor; reserved by hardware and can't be used) 0008 kernel code (ring 0) 0010 kernel data (ring 0) 0018 user code (ring 3) 0020 user data (ring 3) 0028 TSS (hardware defined structure) Isolation between processes would then be done through the page mapping mechanism. This is kind of hard to wrap your head around at first, but makes it much easier to allocate memory. One big problem with using segments instead of fixed-size pages is that when you need to free them in some random order, memory becomes fragmented. But you can also have segments in addition to paging, which is helpful for transitioning from/to 16-bit mode. Hope that helps at least a little bit!
- bogomipz 5y ago
  Sorry "remapping" was a poor choice of words on my part. What I meant was the code segment register gets reloaded to some offset in that GDT that has those protection bits set to 0. According to your flat address space map above that would be at offsets 0008 and 0010. That makes a lot of sense, thanks.