Query regarding x86_64 purgatory and IA32-e compatibility mode

[Vivek Goyal <vgoyal@redhat.com>]

Hi Eric,

I am reading up x86_64 purgatory code to understand how transition to
32bit protected mode happens.

My understanding is that we enter in purgatory_start (setup-x86_64.S).
Then we jump to entry64 in entry64.S.

We run following code in arch/x86_64/entry64.S

        movq    $stack_init, %rsp
        pushq   $0x10 /* CS */
        pushq   $new_cs_exit
        lretq
new_cs_exit:

        /* Load the registers */
        movq    rax(%rip), %rax
        movq    rbx(%rip), %rbx
        movq    rcx(%rip), %rcx
        movq    rdx(%rip), %rdx
        movq    rsi(%rip), %rsi
        movq    rdi(%rip), %rdi
        movq    rsp(%rip), %rsp
        movq    rbp(%rip), %rbp
        movq    r8(%rip), %r8
        movq    r9(%rip), %r9
        movq    r10(%rip), %r10
        movq    r11(%rip), %r11
        movq    r12(%rip), %r12
        movq    r13(%rip), %r13
        movq    r14(%rip), %r14
        movq    r15(%rip), %r15

Will above lretq call not switch us in compatibility mode (from 64bit
mode)? We have taken a long jump and our new CS seems to have L bit 
0.

Following is our gdt.

gdt:    /* 0x00 unusable segment 
         * 0x08 unused
         * so use them as the gdt ptr
         */
        .word   gdt_end - gdt - 1
        .quad   gdt
        .word   0, 0, 0

        /* 0x10 4GB flat code segment */
        .word   0xFFFF, 0x0000, 0x9A00, 0x00AF

        /* 0x18 4GB flat data segment */
        .word   0xFFFF, 0x0000, 0x9200, 0x00CF
gdt_end:

I see that bit 21  in second doubleword is 0. IIUC, that means that we
will switch to compatibility mode. If yes, we are still continuing to
use 64bit instructions and continue to access registers (rip, r8-15)
which are available in 64bit mode only. Is this correct? How does this
work?

Thanks
Vivek

_______________________________________________
kexec mailing list
kexec@lists.infradead.org
http://lists.infradead.org/mailman/listinfo/kexec