operating system - Software initialization code at 0xFFFFFFF0H

Question

Intel says after reset the processor is placed in real mode and the software initialization code starts at 0xFFFFFFF0H. My questions:

• If processor is in real-mode how can it acess the memory > 1MB (0xFFFFFFF0H)

• How this happens or what happens when RAM in < 4GB ( say 2GB)

• If the BIOS is mapped at 0x000FFFFFH why processor starts executing at 0xFFFFFFF0H

Please help me with these questions. Thanks.

Answer 1

I finally found the answer in the Coreboot documentation:

Whenever an x86 CPU wakes up after reset, it does it in Real Mode. This mode is limited to 1MiB address space and 64k offsets and the reset vector of the original 8086/88 was located at 0xFFFF0.

As there was no change even if we run current processors like P3, these newer CPUs also feels like they where start at 0xF000:0xFFF0 after a reset. But they do not. The base of the code segment register is 0xFFFF0000 after reset, so the CPU generates a physical address of 0xFFFFFFF0 to the chipset. And the chipset is responsible to forward this area to the boot ROM. Its confusing: The CPU "thinks" it runs code at 0xF000:0xFFF0 but instead it uses code at 0xFFFFFFF0. The developers must have been tanked up when they realised this design into silicon.

So it seems the Intel documentation talks about the physical address as used "on the wire", i.e. when accessing the real bus. And this is independent of the CPU mode (the bus doesn't know or care about a CPU mode, it's the CPUs duty to translate these things).