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But putting the memory manager in the boot loader isn?t practical. We need to have the processor seeing address 0xC0000000 as 0x100000 without enabling paging. Later, we want to enable paging and avoid relocating the kernel. We can do this by fiddling the base addresses of the kernel?s code and data. Remember that the processor forms physical addresses by adding the virtual address (e.g. 0xC0000000) to the segment?s base address and sending that to the address bus (or, if paging is enabled, to the MMU). So as long as our address space is continuous (that is, there?s no gaps or jumps) we can get the processor to associate any virtual address with any physical address. In our example before, with the kernel located at 0xC0000000 but loaded at 0x100000, we need a segment base address which turns 0xC0000000 into 0x1000000; that is, 0xC0000000 + base = 0x1000000. This is easy: our segment base address needs to be 0x41000000. When we refer to a global variable at address 0xC0001000, the CPU adds 0x41000000 and gets the address 0x1001000. This happens to be where the boot loader loaded part of our kernel, and everyone?s happy. If we later enable paging, and map address 0xC0000000 to 0x1000000 (and use a segment with base address zero), the same addresses will continue to be used.

srg wrote: Anyway, please correct me if I'm wrong but is the maths in this relying on a wrap arround to get 0xC0001000 + 0x41000000 = 0x1001000. If so, what happends on a AMD64 system, would this all have been done in pmode then you switch to long mode, or doesn't it matter? It just sounds like another DOS and A20 Gate type problem if you get what I mean.