shmalloc.asm

来自「ART OF Assembly Language Programming, 很不」· 汇编 代码 · 共 819 行 · 第 1/2 页

; is the first and last item), but this code handles that case without any
; problems.

Free1st:	mov	ax, [bx].Next
		mov	ResidentSeg:AllocatedList, ax
		jmp	AddFree

; If we're removing the last guy in the chain, simply set the next field
; of the previous node in the list to NULL.

FreeLast:	mov	di, [bx].Prev
		mov	[di].Next, NULL

; Okay, now we've got to put the freed block onto the free block list.
; The free block list is sorted according to address.  We have to search
; for the first free block whose address is greater than the block we've
; just freed and insert the new free block before that one.  If the two
; blocks are adjacent, then we've got to merge them into a single free
; block.  Also, if the block before is adjacent, we must merge it as
; well.  This will coalesce all free blocks on the free list so there
; are as few free blocks as possible and those blocks are as large as
; possible.

AddFree:	mov	ax, ResidentSeg:FreeList
		test	ax, ax			;Empty list?
		jne	SrchPosn

; If the list is empty, stick this guy on as the only entry.

		mov	ResidentSeg:FreeList, bx
		mov	[bx].Next, NULL
		mov	[bx].Prev, NULL
		jmp	FreeDone

; If the free list is not empty, search for the position of this block
; in the free list:

SrchPosn:	mov	di, ax
		cmp	bx, di
		jb	FoundPosn
		mov	ax, [di].Next
		test	ax, ax			;At end of list?
		jne	SrchPosn

; If we fall down here, the free block belongs at the end of the list.
; See if we need to merge the new block with the old one.

		mov	ax, di
		add	ax, [di].BlkSize	;Compute address of 1st byte
		add	ax, 8			; after this block.
		cmp	ax, bx
		je	MergeLast

; Okay, just add the free block to the end of the list.

		mov	[di].Next, bx
		mov	[bx].Prev, di
		mov	[bx].Next, NULL
		jmp	FreeDone

; Merge the freed block with the block DI points at.

MergeLast:	mov	ax, [di].BlkSize
		add	ax, [bx].BlkSize
		add	ax, 8
		mov	[di].BlkSize, ax
		jmp	FreeDone

; If we found a free block before which we are supposed to insert
; the current free block, drop down here and handle it.

FoundPosn:      mov	ax, bx			;Compute the address of the
		add	ax, [bx].BlkSize	; next block in memory.
		add	ax, 8
		cmp	ax, di			;Equal to this block?
		jne	DontMerge

; The next free block is adjacent to the one we're freeing, so just
; merge the two.

		mov	ax, [di].BlkSize	;Merge the sizes together.
		add	ax, 8
		add	[bx].BlkSize, ax
		mov	ax, [di].Next		;Tweak the links.
		mov	[bx].Next, ax
		mov	ax, [di].Prev
		mov	[bx].Prev, ax
		jmp	TryMergeB4

; If the blocks are not adjacent, just link them together here.

DontMerge:	mov	ax, [di].Prev
		mov	[di].Prev, bx
		mov	[bx].Prev, ax
		mov	[bx].Next, di

; Now, see if we can merge the current free block with the previous free blk.

TryMergeB4:	mov	di, [bx].Prev
		mov	ax, di
		add	ax, [di].BlkSize
		add	ax, 8
		cmp	ax, bx
		je	CanMerge
		pop	bx
		pop	di			;Nothing allocated, just
		pop	ds			; return to caller.
		iret

; If we can merge the previous and current free blocks, do that here:

CanMerge:	mov	ax, [bx].Next
		mov	[di].Next, ax
		mov	ax, [bx].BlkSize
		add	ax, 8
		add	[di].BlkSize, ax
		pop	bx
		pop	di
		pop	ds
		iret

		assume	ds:nothing
		assume	bx:nothing
		assume	di:nothing

; Here's where we handle the shared memory initializatin (SHMINIT) function.
; All we got to do is create a single block on the free list (which is all
; available memory), empty out the allocated list, and then zero out all
; shared memory.

Tryshminit:	cmp	al, 13h
		jne	TryShmAttach

; Reset the memory allocation area to contain a single, free, block of
; memory whose size is 0FFF8h (need to reserve eight bytes for the block's
; data structure).

		push	es
		push	di
		push	cx

		mov	ax, SharedMemory	;Zero out the shared
		mov	es, ax			; memory segment.
		mov	cx, 32768
		xor	ax, ax
		mov	di, ax
	rep	stosw


; Note: the commented out lines below are unnecessary since the code above
; has already zeroed out the entire shared memory segment.
; Note: we cannot put the first record at offset zero because offset zero
; is the special value for the NULL pointer.  We'll use 4 instead.

		mov	di, 4
;		mov	es:[di].Region.Key, 0		;Key is arbitrary.
;		mov	es:[di].Region.Next, 0		;No other entries.
;		mov	es:[di].Region.Prev, 0		; Ditto.
		mov	es:[di].Region.BlkSize, 0FFF8h	;Rest of segment.
		mov	ResidentSeg:FreeList, di

		pop	cx
		pop	di
		pop	es
		mov	ax, 0				;Return no error.
		iret


; Handle the SHMATTACH function here.  On entry, DX contains a key number.
; Search for an allocated block with that key number and return a pointer
; to that block (if found) in ES:DI.  Return an error code (AX=3) if we
; cannot find the block.

TryShmAttach:	cmp	al, 14h			;Attach opcode.
		jne	IllegalOp
		mov	ax, SharedMemory
		mov	es, ax

		mov	di, ResidentSeg:AllocatedList
FindOurs:	cmp	dx, es:[di].Region.Key
		je	FoundOurs
		mov	di, es:[di].Region.Next
		test	di, di
		jne	FoundOurs
		mov	ax, 3			;Can't find the key.
		iret

FoundOurs:	add	di, 8			;Point at actual data.
		mov	ax, 0			;No error.
		iret


; They called us with an illegal subfunction value.  Try to do as little
; damage as possible.

IllegalOp:	mov	ax, 0		;Who knows what they were thinking?
		iret
MyInt2F		endp
		assume	ds:nothing
ResidentSeg	ends


; Here's the segment that will actually hold the shared data.

SharedMemory	segment	para public 'Shared'
		db	0FFFFh dup (?)
SharedMemory	ends






cseg		segment	para public 'code'
		assume	cs:cseg, ds:ResidentSeg

; SeeIfPresent-	Checks to see if our TSR is already present in memory.
;		Sets the zero flag if it is, clears the zero flag if
;		it is not.

SeeIfPresent	proc	near
		push	es
		push	ds
		push	di
		mov	cx, 0ffh		;Start with ID 0FFh.
IDLoop:		mov	ah, cl
		push	cx
		mov	al, 0			;Verify presence call.
		int	2Fh
		pop	cx
		cmp	al, 0			;Present in memory?
		je	TryNext
		strcmpl
		byte	"Dynamic Shared Memory TSR",0
		je	Success

TryNext:	dec	cl			;Test USER IDs of 80h..FFh
		js	IDLoop
		cmp	cx, 0			;Clear zero flag.
Success:	pop	di
		pop	ds
		pop	es
		ret
SeeIfPresent	endp



; FindID-	Determines the first (well, last actually) TSR ID available
;		in the multiplex interrupt chain.  Returns this value in
;		the CL register.
;
;		Returns the zero flag set if it locates an empty slot.
;		Returns the zero flag clear if failure.

FindID		proc	near
		push	es
		push	ds
		push	di

		mov	cx, 0ffh		;Start with ID 0FFh.
IDLoop:		mov	ah, cl
		push	cx
		mov	al, 0			;Verify presence call.
		int	2Fh
		pop	cx
		cmp	al, 0			;Present in memory?
		je	Success
		dec	cl			;Test USER IDs of 80h..FFh
		js	IDLoop
		xor	cx, cx
		cmp	cx, 1			;Clear zero flag
Success:	pop	di
		pop	ds
		pop	es
		ret
FindID		endp



Main		proc
		meminit

		mov	ax, ResidentSeg
		mov	ds, ax

		mov	ah, 62h			;Get this program's PSP
		int	21h			; value.
		mov	PSP, bx

; Before we do anything else, we need to check the command line
; parameters.  If there is one, and it is the word "REMOVE", then remove
; the resident copy from memory using the multiplex (2Fh) interrupt.

		argc
		cmp	cx, 1			;Must have 0 or 1 parms.
		jb	TstPresent
		je	DoRemove
Usage:		print
		byte	"Usage:",cr,lf
		byte	"       shmalloc",cr,lf
		byte	"or     shmalloc REMOVE",cr,lf,0
		ExitPgm


; Check for the REMOVE command.

DoRemove:	mov	ax, 1
		argv
		stricmpl
		byte	"REMOVE",0
		jne	Usage

		call	SeeIfPresent
		je	RemoveIt
		print
		byte	"TSR is not present in memory, cannot remove"
		byte	cr,lf,0
		ExitPgm

RemoveIt:	mov	MyTSRID, cl
		printf
		byte	"Removing TSR (ID #%d) from memory...",0
		dword	MyTSRID

		mov	ah, cl
		mov	al, 1			;Remove cmd, ah contains ID
		int	2Fh
		cmp	al, 1			;Succeed?
		je	RmvFailure
		print
		byte	"removed.",cr,lf,0
		ExitPgm

RmvFailure:	print
		byte	cr,lf
		byte	"Could not remove TSR from memory.",cr,lf
		byte	"Try removing other TSRs in the reverse order "
		byte	"you installed them.",cr,lf,0
		ExitPgm



; Okay, see if the TSR is already in memory.  If so, abort the
; installation process.

TstPresent:     call	SeeIfPresent
		jne	GetTSRID
		print
		byte	"TSR is already present in memory.",cr,lf
		byte	"Aborting installation process",cr,lf,0
		ExitPgm


; Get an ID for our TSR and save it away.

GetTSRID:	call	FindID
		je	GetFileName
		print
		byte	"Too many resident TSRs, cannot install",cr,lf,0
		ExitPgm


; Things look cool so far, so install the interrupts

GetFileName:	mov	MyTSRID, cl
		print
		byte	"Installing interrupts...",0


; Patch into the INT 2Fh interrupt chain.

		cli				;Turn off interrupts!
		mov	ax, 0
		mov	es, ax
		mov	ax, es:[2Fh*4]
		mov	word ptr OldInt2F, ax
		mov     ax, es:[2Fh*4 + 2]
		mov	word ptr OldInt2F+2, ax
		mov	es:[2Fh*4], offset MyInt2F
		mov	es:[2Fh*4+2], seg ResidentSeg
		sti				;Okay, ints back on.

; We're hooked up, the only thing that remains is to initialize the shared
; memory segment and then terminate and stay resident.

		printf
		byte	"Installed, TSR ID #%d.",cr,lf,0
		dword	MyTSRID

		mov	ah, MyTSRID		;Initialization call.
		mov	al, 13h
		int	2Fh

		mov	dx, EndResident		;Compute size of program.
		sub	dx, PSP
		mov	ax, 3100h		;DOS TSR command.
		int	21h
Main		endp
cseg		ends

sseg		segment	para stack 'stack'
stk		db	256 dup (?)
sseg		ends

zzzzzzseg	segment	para public 'zzzzzz'
LastBytes	db	16 dup (?)
zzzzzzseg	ends
		end	Main