how to use LDT

[ckzippo]

I want to use LDT. so i define a new selector in GDT.

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   mov	$gdt, %eax
	addl $FIRST_LDT_SELECTOR, %eax
	movw	$(ldt0end - ldt0 - 1), (%eax)
	movl	 $ldt0, %ebx
	movw	%bx, 2(%eax)
	shr    $16, %ebx
	movb	%bl, 4(%eax)
	movb	%bh, 7(%eax)
	movb	$0x82,5(%eax)

the code above is to set the descriptor in GDT.
and ldt is defined as below:

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ldt0:
	.quad 0x0000000000000000	# first is not in use
	.quad 0x00cf981000c1ffff
ldt0end:

then , i use the following:

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  movl	$FIRST_LDT_SELECTOR, %eax
	lldt	%ax                      # error
   jmp   $0x0c, $0

i use qemu to debug, and it crashes when lldt is excuted.
i'm confused with this problem several days.

Any help will be great appreciated.Thank you.

[Combuster]

I suggest that you try bochs, it provides much more detailed output when something goes wrong. It also has a debugger version where you can actually inspect the GDT and LDT in detail at the location of the crash.

As far as the code is concerned, it occurs to me that byte 6 of the GDT entry is left undefined.

[ckzippo]

I suggest that you try bochs, it provides much more detailed output when something goes wrong. It also has a debugger version where you can actually inspect the GDT and LDT in detail at the location of the crash.

As far as the code is concerned, it occurs to me that byte 6 of the GDT entry is left undefined.

byte 6 is all 0.
i use

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movw $0x82,5(%eax)

rather

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movb $0x82,5(%eax)

it generates the same fault, GP.
anyway,thank you.i will try bochs.

[amd64pager]

byte 6 is all 0.
i use

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movw $0x82,5(%eax)

rather

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movb $0x82,5(%eax)

it generates the same fault, GP.
anyway,thank you.i will try bochs.

Try

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movb $0x82,5(%eax)
movb $0x00,6(%eax)

[ckzippo]

byte 6 is all 0.
i use

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movw $0x82,5(%eax)

rather

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movb $0x82,5(%eax)

it generates the same fault, GP.
anyway,thank you.i will try bochs.

Try

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movb $0x82,5(%eax)
movb $0x00,6(%eax)

yes,i tested.But the same,GP.
thanks.

[amd64pager]

Instead of using this:

and ldt is defined as below:

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ldt0:
	.quad 0x0000000000000000	# first is not in use
	.quad 0x00cf981000c1ffff
ldt0end:

maybe you could break it down into bytes and words.

EDIT:I think my old bootsector(an year ago,I think) must have failed because of this(triple faulted)

[ckzippo]

Instead of using this:

and ldt is defined as below:

Code: Select all

ldt0:
	.quad 0x0000000000000000	# first is not in use
	.quad 0x00cf981000c1ffff
ldt0end:

maybe you could break it down into bytes and words.

EDIT:I think my old bootsector(an year ago,I think) must have failed because of this(triple faulted)

Does this matter? i checked the memory in qemu. i find whether break the quad down to bytes and words or not,the organisation
in memory is the same.
Also, i try this:

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.word 0xffff,0x00c1,0x9810,0x00cf

and this:

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.byte 0xff,0xff,0xc1,0x00
.byte 0x10,0x98,0xcf,0x00

the result is also GP.
Thank you.

[amd64pager]

Could you tell me what is your descriptors base,limit,etc.?
Maybe I could rewrite it for you.

[rdos]

Compared with my own create ldt code, and the code to create the ldt-descriptor seems to be ok. What I wonder is what the descriptor index is, and what the GDT-limit is? Since you create the descriptor within a flat memory model, you might succeed with doing that without faults even if you are outside of the GDT-limit. That would explain the GPF. The load of the LDT should not fail when offset 5 is 0x82.

[ckzippo]

Could you tell me what is your descriptors base,limit,etc.?
Maybe I could rewrite it for you.

i defined GDT as :

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struct segdesc_s gdt[GDT_SIZE]

and marco GDT_SIZE = 100;
the selector of LDT is FIRST_LDT_SELECTOR,whose value is 0x50.
the base of GDT is gdt, and the limit is GDT_SIZE * 8 -1
i think i set GDT right ,because when i use other selectors in GDT,it is OK.

the base of LDT is ldt0, which is defined as:

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ldt0:
     .quad 0x0000000000000000   # first is not in use
     .quad 0x00cf9a000000ffff       #selector is 0x0C,it's local code segment descriptor
ldt0end:

the limit of LDT is:
ldt0end - ldt0 - 1
then i use a jmp to jmp to LDT.

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movl $FIRST_LDT_SELECTOR, %eax
lldt  %ax                           #error
jmp $0x0c,$task0
task0:

the error occured lldt %ax.

i write exception handlers in my kernel.
i set 13th exception handler to print "General Protection"
i set 16th exception handler to print "FPU Floating Point Error(Math Fault)"

it first print Geneal Protection, and after a while , it print FPU Floating point Error

thank you.

[ckzippo]

Compared with my own create ldt code, and the code to create the ldt-descriptor seems to be ok. What I wonder is what the descriptor index is, and what the GDT-limit is? Since you create the descriptor within a flat memory model, you might succeed with doing that without faults even if you are outside of the GDT-limit. That would explain the GPF. The load of the LDT should not fail when offset 5 is 0x82.

the selector is 0x50, and GDT limit is 100*8 -1 . And it's not outside the GDT table.
Thank you.

[amd64pager]

If the source operand is 0, the LDTR is marked invalid and all references to descriptors in the LDT (except by the LAR, VERR, VERW or LSL instructions) cause a general protection excep-tion (#GP).

-from http://faydoc.tripod.com/cpu/lldt.htm

[amd64pager]

Also , try:

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lldt $0x50

[Combuster]

anyway,thank you.i will try bochs.

Have you yet?

[ckzippo]

Also , try:

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lldt $0x50

we can not use

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lldt $0x50

according to the Intel Manual,
we should use

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LLDT r/m16

can not use an imm like $0x50

and this indeed cause error "operand type mismatch for lldt"

thank you for your advice.