"mov rdi,0x000B8000" (was: What does your OS look like?)

[jal]

It was later changed to this:

Code: Select all

	mov rdi, 0x00000000000B8000
	mov rax, 0x0720075507500743	; 'CPU '
	stosq

Wouldn't there be a shorter way (in bytes) to assign b8000 to rdi? Like first xor-ing rdi, then assigning it to edi? or to di, then shifting, or whatever?


JAL

[IanSeyler]

It was later changed to this:

Code: Select all

	mov rdi, 0x00000000000B8000
	mov rax, 0x0720075507500743	; 'CPU '
	stosq

Wouldn't there be a shorter way (in bytes) to assign b8000 to rdi? Like first xor-ing rdi, then assigning it to edi? or to di, then shifting, or whatever?


JAL

Yes. The above code is 10 bytes. The below code is 7 bytes. DI with a shift is 9 bytes.

Code: Select all

	xor edi, edi			; Automatically clears the high 32 bits as well
	mov edi, 0x000B8000

[Firestryke31]

For some reason I am under the impression that a simple mov edi, 0x000B8000 would also clear the high half of rdi. It's probably me misremembering something I read here, though.

[Selenic]

For some reason I am under the impression that a simple mov edi, 0x000B8000 would also clear the high half of rdi. It's probably me misremembering something I read here, though.

No, that's what happens. It was included by AMD when designing long mode in order to reduce partial register dependencies, I think (but 8- and 16-bit ones don't clear anything, for backwards-compatibility reasons)

[IanSeyler]

I verified it in the latest version of bochs. "mov edi, 0x000b8000" clears the higher 32-bits. I was unsure if the mov instruction did that but now I know.

[Love4Boobies]

At the price of using an extra register,

Code: Select all

xor ebx, ebx
mov bh, 0b8h
mov edi, ebx

results in 6 bytes (if you really must use RDI ).

[Firestryke31]

except that makes edi 0x0000B800, not 0x000B8000 (it's shifted down a nibble for those too lazy to count digits).

[Gigasoft]

And mov edi,0b8000h is only 5 bytes anyway.

[Love4Boobies]

Though that doesn't clear the high dword so it doesn't work.

[Firestryke31]

Though that doesn't clear the high dword so it doesn't work.

I verified it in the latest version of bochs. "mov edi, 0x000b8000" clears the higher 32-bits.

It seems as though it actually does.

[Brendan]

Hi,

The instruction "mov rdi, 0x00000000000B8000" should be assembled into whatever opcode the assembler thinks is best. Unfortunately the optimisers in most assemblers aren't very good, so...

"mov rdi, 0x00000000000B8000" = 10 bytes

"mov edi, 0x000B8000" = 5 bytes and does exactly the same thing (the CPU does clear the high dword)

"xor ebx, ebx; mov bh, 0b8h; mov edi, ebx" = wrong/buggy (0x0000B800 != 0x000B8000)

mov rdi,[address_of_video_memory] = 8 bytes for the instruction plus 8 bytes of data; and even though it's longer and slower it's the correct way of doing it (because it won't break later when you decide to use a double buffer, or when you start using VBE and LFB for setting a text mode, or when you're using graphics mode with a buffer in RAM for "legacy" stuff).


Cheers,

Brendan

[Love4Boobies]

"xor ebx, ebx; mov bh, 0b8h; mov edi, ebx" = wrong/buggy (0x0000B800 != 0x000B8000)

Yes, we've been through that, I didn't notice the address, sue me!

[jal]

Just because I'm a lazy bum who's never dabbled in 64 bit before: is the high dword of rXX cleared or is it sign extended? I'd expect the latter.


JAL

[Gigasoft]

The high dword is always cleared.