kolibrios-fun/kernel/trunk/core/slab.inc

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;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; ;;
;; Copyright (C) KolibriOS team 2013-2015. All rights reserved. ;;
;; Distributed under terms of the GNU General Public License ;;
;; ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
$Revision$
; Memory management for slab structures.
; The allocator meets special requirements:
; * memory blocks are properly aligned
; * memory blocks do not cross page boundary
; The allocator manages fixed-size blocks.
; Thus, the specific allocator works as follows:
; allocate one page, split into blocks, maintain the single-linked
; list of all free blocks in each page.
; Note: size must be a multiple of required alignment.
; Data for one pool: dd pointer to the first page, MUTEX lock.
; Allocator for fixed-size blocks: allocate a block.
; [ebx-4] = pointer to the first page, ebx = pointer to MUTEX structure.
proc slab_alloc
push edi ; save used register to be stdcall
virtual at esp
dd ? ; saved edi
dd ? ; return address
.size dd ?
end virtual
; 1. Take the lock.
mov ecx, ebx
call mutex_lock
; 2. Find the first allocated page with a free block, if any.
; 2a. Initialize for the loop.
mov edx, ebx
.pageloop:
; 2b. Get the next page, keeping the current in eax.
mov eax, edx
mov edx, [edx-4]
; 2c. If there is no next page, we're out of luck; go to 4.
test edx, edx
jz .newpage
add edx, 0x1000
@@:
; 2d. Get the pointer to the first free block on this page.
; If there is no free block, continue to 2b.
mov eax, [edx-8]
test eax, eax
jz .pageloop
; 2e. Get the pointer to the next free block.
mov ecx, [eax]
; 2f. Update the pointer to the first free block from eax to ecx.
; Normally [edx-8] still contains eax, if so, atomically set it to ecx
; and proceed to 3.
; However, the price of simplicity of slab_free (in particular, it doesn't take
; the lock) is that [edx-8] could (rarely) be changed while we processed steps
; 2d+2e. If so, return to 2d and retry.
lock cmpxchg [edx-8], ecx
jnz @b
.return:
; 3. Release the lock taken in step 1 and return.
push eax
mov ecx, ebx
call mutex_unlock
pop eax
pop edi ; restore used register to be stdcall
ret 4
.newpage:
; 4. Allocate a new page.
push eax
stdcall kernel_alloc, 0x1000
pop edx
; If failed, say something to the debug board and return zero.
test eax, eax
jz .nomemory
; 5. Add the new page to the tail of list of allocated pages.
mov [edx-4], eax
; 6. Initialize two service dwords in the end of page:
; first free block is (start of page) + (block size)
; (we will return first block at (start of page), so consider it allocated),
; no next page.
mov edx, eax
lea edi, [eax+0x1000-8]
add edx, [.size]
mov [edi], edx
and dword [edi+4], 0
; 7. All blocks starting from edx are free; join them in a single-linked list.
@@:
mov ecx, edx
add edx, [.size]
mov [ecx], edx
cmp edx, edi
jbe @b
sub ecx, [.size]
and dword [ecx], 0
; 8. Return (start of page).
jmp .return
.nomemory:
dbgstr 'no memory for slab allocation'
xor eax, eax
jmp .return
endp
; Allocator for fixed-size blocks: free a block.
proc slab_free
push ecx edx
virtual at esp
rd 2 ; saved registers
dd ? ; return address
.block dd ?
end virtual
; Insert the given block to the head of free blocks in this page.
mov ecx, [.block]
mov edx, ecx
or edx, 0xFFF
@@:
mov eax, [edx+1-8]
mov [ecx], eax
lock cmpxchg [edx+1-8], ecx
jnz @b
pop edx ecx
ret 4
endp