kolibrios/kernel/branches/kolibri-process/fs/ext2/resource.inc

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;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; ;;
;; Contains common resource allocation + freeing code. ;;
;; ;;
;; Copyright (C) KolibriOS team 2004-2013. All rights reserved. ;;
;; Distributed under the terms of the new BSD license. ;;
;; ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;---------------------------------------------------------------------
; Frees a resource (block/inode).
; Input: eax = resource ID.
; edi = function pointer of ext2_bg_*_bitmap form, to
; get bitmap of resource.
; ecx = 0, block; 1, inode.
; ebp = pointer to EXTFS.
; Output: Block marked as free in block group.
; eax = error code.
;---------------------------------------------------------------------
ext2_resource_free:
push ebx edx esi
; Get block group.
sub eax, [ebp + EXTFS.superblock + EXT2_SB_STRUC.first_data_block]
xor edx, edx
div [ebp + EXTFS.superblock + EXT2_SB_STRUC.blocks_per_group]
push eax edx
call edi
test eax, eax
jz .fail
mov esi, eax
; Read the bitmap.
mov eax, ebx
mov edx, eax
mov ebx, [ebp + EXTFS.ext2_save_block]
call ext2_block_read
test eax, eax
jnz .fail
pop eax
; Mark bit free.
call bitmap_clear_bit
test eax, eax
jz @F
; No need to save anything.
xor eax, eax
add esp, 4
jmp .return
@@:
mov eax, edx
mov ebx, [ebp + EXTFS.ext2_save_block]
call ext2_block_write
test eax, eax
jnz .fail
; Read the descriptor.
mov eax, [esp]
call ext2_bg_read_desc
test eax, eax
jz .fail_bg_desc_read
lea eax, [eax + EXT2_BLOCK_GROUP_DESC.free_blocks_count]
shl ecx, 1
add eax, ecx
inc word[eax]
lea eax, [ebp + EXTFS.superblock + EXT2_SB_STRUC.free_block_count]
shl ecx, 1
add eax, ecx
inc dword[eax]
pop eax
call ext2_bg_write_desc
.return:
pop esi edx ebx
ret
.fail:
add esp, 4
.fail_bg_desc_read:
add esp, 4
xor eax, eax
not eax
jmp .return
;---------------------------------------------------------------------
; Allocates a resource.
; Input: eax = inode ID for "preference".
; ebp = pointer to EXTFS.
; [esp + 4], func pointer to ext2_bg_*_bitmap
; [esp + 8], pointer to free_*_count in SB.
; [esp + 12], *_per_group
; [esp + 16], offset to free_*_count in bg descriptor.
; [esp + 20], *_count
; Output: Resource marked as set in block group.
; eax = error code.
; ebx = resource ID.
;---------------------------------------------------------------------
ext2_resource_alloc:
; Block allocation is a pretty serious area, since bad allocation
; can lead to fragmentation. Thus, the best way to allocate that
; comes to mind is to allocate around an inode as much as possible.
; On the other hand, this isn't about a single inode/file/directory,
; and focusing just around the preferred inode would lead to
; congestion. Thus, after much thought, the chosen allocation algorithm
; is to search forward, then backward.
push ecx edx esi edi
cmp dword[esp + 16 + 8], 0
jnz @F
; No free blocks.
xor eax, eax
not eax
pop edi esi edx ecx
ret 20
@@:
; Calculate which block group the preferred inode belongs to.
dec eax
xor edx, edx
; EAX = block group.
div [ebp + EXTFS.superblock + EXT2_SB_STRUC.inodes_per_group]
push eax
push eax
mov edi, .forward
.test_block_group:
call dword[esp + 16 + 8 + 4]
test eax, eax
jz .fail
mov esi, eax
mov eax, ebx
mov edx, eax
mov ebx, [ebp + EXTFS.ext2_save_block]
call ext2_block_read
test eax, eax
jnz .fail
mov ecx, [esp + 16 + 8 + 12]
call ext2_find_free_bit
cmp eax, 0xFFFFFFFF
jne @F
mov eax, edi
jmp eax
@@:
mov ecx, eax
mov eax, edx
mov ebx, [ebp + EXTFS.ext2_save_block]
call ext2_block_write
test eax, eax
jnz .fail
; ecx: the index of the matched entry.
; [esp]: block group where we found.
; [esp + 4]: starting block group.
; esi: block group descriptor.
mov eax, [esp] ; Index of block group in which we found.
mul dword[esp + 16 + 8 + 12]
add eax, ecx
mov ebx, eax
mov eax, [esp + 16 + 8 + 8]
dec dword[eax]
mov eax, esi
add eax, [esp + 16 + 8 + 16]
dec word[eax]
pop eax
call ext2_bg_write_desc
add esp, 4
jmp .return
; Continue forward.
.forward:
inc dword[esp]
mov eax, [esp]
mul dword[esp + 16 + 8 + 12]
cmp eax, [esp + 16 + 8 + 20]
jbe @F
; We need to go backward.
mov eax, [esp + 4]
mov [esp], eax
mov edi, .backward
jmp .backward
@@:
mov eax, [esp]
jmp .test_block_group
; Continue backward.
.backward:
cmp dword[esp], 0
je .fail
dec dword[esp]
mov eax, [esp]
jmp .test_block_group
.return:
pop edi esi edx ecx
ret 20
.fail:
add esp, 8
xor eax, eax
not eax
jmp .return