These are the three steps that need to be taken in order to successfully activate Long Mode. Error reporting is sparse as most real hardware for developing an OS is 64 bit but probably still a good idea to implement this verification procedure.

I did a little tweaking of some of the examples that can be found on the web either in whole or in part and believe 74 bytes might be as tight as it can get.

    PageZero    equ 0xb800
        ID  equ 0b100000

; AMD 24592 Rev 3.14 2007 Feature detection. Section 3.6 pg 74
; ------------------------------------------------------------

  ; A: Test if CPU supports CPUID

    pushfd          ; Save a current copy of flags

    ; Create a simplified pointer that becomes essentially SS:BX

    push    ss
    mov     bx, sp      ; BX will point to bits 23-16 of EFLAGS
    pop     ds

    ; Read flags, toggle (ID) and write back to flags.

    pushfd                  ; Get another copy of EFLAGS 
    or      byte [bx], ID   ; Toggle bit 21 (ID) [bit 5 of BX]
    mov     al, [bx]        ; Save a copy
    popfd                   ; Write value back to EFLAGS

    ; Read flags again and if ID remains inverted, then processor supports CPUID.

    pop     dx              ; DX = Bits 15 - 0 of EFLAGS
    pop     dx              ; DX = Bits 16 - 31
    popfd                   ; Restore flags to original values
    xor     al, dl          ; Bit 4 of AL & DL is (ID)
    jz      TstExt

; As this error is very improbable, near center bottom there will be an uppercase "E"
; yellow on red, with two red flashing bars on each side.

FncErr: mov di, (22*80+39)*2 ; Point to center position of 21th row in video
        mov ax, PageZero     ; Page 0 of 80x26x16 (Mode 3)
        mov es, ax
        mov eax, 0x2e4684b1 ; This makes the desired character combination work

        jmp     $ - 1       ; Just hang.

  ; B: Are extended functions supported

    mov     eax, 0x80000000
    push    eax
    cpuid           ; Extended function limit
    pop     edx
    cmp     eax, edx
    jb      FncErr

  ; C: Does processor support Long Mode.

    mov     al, 1       ; Set EAX = 0x80000001
    bt      edx, 29     ; 64 bit available if bit is on.
    jnc     FncErr

1 Answer 1


Documentation for this flag reads:

If a software procedure can set and clear this flag, the processor executing the procedure supports the CPUID instruction

This clearly means that the flag does not have a defined value. It could be OFF or ON at any time!

  • Where your comment says that you "Toggle bit 21 (ID)", the instruction actually sets the bit. I don't think that would be too useful since you don't use the pre-existing value of this bit.

  • When you exclusive or XOR the registers AL and DL, you should prudently mask away the 7 remaining bits.

All the forementioned boils down to the fact that the final test (notice the typo in the comment; should be bit 5)

xor     al, dl          ; Bit 4 of AL & DL is (ID)
jz      TstExt

will always check out fine.

Although your method to get a DS:BX pointer is quite ingenious, I prefer using [bp-2] as you can see in below code.

    mov     bp, sp
    mov     al, [bp-2]        ;Pre-existing value
    xor     byte [bp-2], ID   ;Toggle bit 21
    mov     ah, [bp-2]        ;Hopefully changed value
    and     ax, (ID<<8)+ID    ;Mask off unimportant bits
    cmp     al, ah
    jne     TstExt


I omitted preserving the ID bit as per definition this should not be necessary.

FncErr: mov di, (22*80+39)*2 ; Point to center position of 21th row in video

The address points to the 23rd row.

cmp     eax, edx   <<< EDX=0x80000000
jb      FncErr

Are you sure the service can't report EAX=0x80000000 ?
Perhaps it could be better to bail out with:

jbe     FncErr

Now you're certain that leaf 0x80000001 exists.

Reaching leaf 0x80000001 through mov al, 1 will continue to work for as long as the manufacturer doesn't create some high numbered leafs (0x80000100 and up). You have to ask yourself how future-proof that you want your program to be!

  • \$\begingroup\$ Thanks for the critique and your comments do speak to making the code more predictable. Obviously, it can't be known by the example, but my bootloader sets BP so I would have to essentially create a procedure frame. That being said, I was surprised to find your example is only 2 bytes larger than mine, which at a glance wasn't something I'd expected. \$\endgroup\$
    – Shift_Left
    Nov 11, 2018 at 23:49

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