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【汉化资料】通用解包器制作工具QuickBMS图文教程2

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【汉化资料】通用解包器制作工具QuickBMS图文教程2

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发表于 2009-6-11 20:21 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式

【汉化资料】通用解包器制作工具QuickBMS图文教程2

ok so set up our c:\temp directory bye extracting our file BoneObject.hsp) x, }4 k! Y/ k" t
1. to c:\temp
5 t  q8 p7 a/ C$ q5 q; V8 b2. create a new text document called astro.bms
8 m  L* W1 r/ r2 Y: l7 |- ]! H3. and place the newest version of quickbms in the folder also.: O, m+ H2 n9 p

" O$ h. D( X7 b3 N# X7 nOk so open up BoneObject.hsp in your hex editor and lets take a look at it.
3 \# t! y  K1 m: m  
( q; A/ |. Y2 F! Jgood we have some plain text.+ C  h5 l4 }/ H- O
you will notice I highlighted the first 4 bytes 20 50 53 48 or " PSH" that is a space followed bye P S H.
! A& S) Z7 y& `. E* S$ o/ t* h9 U" Rhmm that seems familiar that is the file extension only backwards. this is know as the idstring
6 m& y1 s& {- C/ Aso up until now you would think to write in bms
; k; f3 t2 w! T0 G- V, R( uget IDSTRING long
7 V4 b# W8 v) o9 P- y6 xthere is nothing wrong with that but there is a better command
* U8 c. l* b  ?/ X: A7 u7 N. |idstring " PSH"8 `/ L  ?6 ~5 s* @) l; ^" o/ w6 B2 I
make sure you include the quotes.
9 R& [# e* B  q2 j! S& W  X/ jso open your bms string and on the first line type# M. q" P; E% Z* v
idstring " PSH"
" n$ ^9 X: I7 U; ?the reason this command is better is it will tell the program not to run if it does not find that string don't try to extract that file.$ X, |& ?$ U# b: d" C
"aka noob proofing it"3 Z* e9 F7 }" V" s3 y7 v; O, ?: `

& z$ E4 c, h- c3 o8 n  p9 r  h9 LOk so now lets look at what we can read I see % _; {! O5 t! ~) W3 S0 l
Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds , Datas\Texture\BoneObject\Toon.bmp , Datas\Texture\BoneObject\Toon_a.bmp , Datas\Texture\BoneObject\Toon_zero.bmp
9 d* T6 B3 ^& P1 J+ H' @6 C  U4 Zso I will assume there are 4 files in this archive.
! ^3 l* p+ h; ]  X8 O. Uwell lets look at the next 4 bytes and see what it is 01 00 00 00 hmm that is equal to 00 00 00 01 or 1 and we have more files in this archive than that so we do not know what this represents/ W+ U: p; T3 |! L, h' m* F! ]: j# n
so lets write that in bms language. Q+ q  d1 h0 Z7 ~2 w: s
get UNK1 long
9 M: f4 v- r, \, Y1 }: [6 zthis saves those 4 bytes as the variable UNK1.- w! V" k" W9 L- @% @1 @

& N  P9 ]" L" l; T4 k9 R$ wok the next 4 bytes are 04 00 00 00 hmm this translates into 00 00 00 04 or 4
$ q/ G; F) z+ ?: C: dhey that is the number of files we counted so lets write that in bms# Y: b9 I0 E) N
get FILES long+ a, w0 a2 {% x, t$ ]
this saves those 4 bytes as the variable FILES.
( v* B, I( b8 l# E# y4 z' W" q
2 O+ t3 ^+ i  d. h) Mthe next 4 bytes are 00 00 00 00 well that is equal to zero so for now I will write that in bms  ~& ]7 l6 A3 g) Q! }
get NULL1 long
+ z* ^# p" l& w5 m# K6 m& f, `this saves those 4 bytes as the variable NULL1
2 E" P6 c6 U7 `/ l3 y( h/ ^% Y
! C, D1 C7 i# W+ B4 N. |( D7 [ok now we have reached the first file name Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds
4 Y& a) K: v( ?) p/ v/ ~5 K3 D% mthis is 0x36 bytes long but wait there was no indicator like the last file that told us how long the name is how do we write this?; T6 q! `, E5 ^( o- C1 Y
well lets look for a pattern  e( W' E+ ~1 o4 X, r& c* T4 d
Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds is 0x36) q  A  G- Z8 C6 \2 p! `1 U1 i
Datas\Texture\BoneObject\Toon.bmp is 0x21
8 W2 {( @+ ^' C2 q9 f& v, Q. PDatas\Texture\BoneObject\Toon_a.bmp is 0x23
+ q. t* z7 i9 K7 E% i' vDatas\Texture\BoneObject\Toon_zero.bmp is 0x26! r- z* a% o( D; I- K4 ]3 t& ~) M1 w
hmm I don't see anything that makes that a pattern.% q- ^! e1 d& R
but I do see all the names are followed bye a lot of zero's. how long is the name + the zeros of each file?1 |2 y: }6 u& d& _; Q
Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds + 0's is 0x80! G  K$ e2 r/ @' S( D8 m& r
Datas\Texture\BoneObject\Toon.bmp + 0's is 0x80
2 k5 R8 R- H" N. Y7 g1 bDatas\Texture\BoneObject\Toon_a.bmp + 0's is 0x803 h; s. h) ~7 z4 S
Datas\Texture\BoneObject\Toon_zero.bmp + 0's is 0x80* c9 C& V6 W! h* o
hey they are all the same size when I include the 0's2 M( w# `5 Y% g
so in bms I would write this as- A6 Y  n6 w8 I. K+ _! L# t
getdstring NAME 0x80
; ]$ G. I& m% ?" u6 b8 u  e7 f1 g# Lthis tells it to grab 0x80 bytes and store the text value of it
/ ?4 c* j4 E% nand as an added feature it will automatically remove trailing 0's 4 E) q9 f/ v5 \" ^

! Z4 w5 o/ t+ c0 @& x( `) ]ok so now we have 0xC bytes before I see the next file name& C( m5 T5 s6 p# Y3 C8 O3 i" W
which is 3 long values& r; S: e$ ?0 @/ g) A/ O, K- |5 x" n
so lets write those in and we will figure out what they represent later.% N5 E! @. n+ u0 q0 D6 f
get UNK2 long, a. |5 C. D" A% [: {: J# _
get UNK3 long
/ K2 U" A, M$ N- y* _, O: Q$ ~+ uget UNK4 long' l3 i& F. R% E0 ?0 Q0 T  Y; Q

( k, X5 k1 Q! W% P2 u3 L8 ook so now we see the name again
. r7 N. d6 T! Q: [* u' ]9 M! Q4 Z! hwe have our pattern so lets write our script based on what we learned' \8 @: `) F6 X- D# @3 @3 @6 }9 n
so it would look like this up until now
6 h* o2 \( ?4 ?" Z
6 f% E7 Z/ k. m, e/ U( w$ WCode:3 E- r0 F+ Z5 y9 w+ j
idstring " PSH"
5 h3 R: m2 C1 f: bget UNK1 long6 `2 t- x& U; ~7 C* h7 @" H
get FILES long; S+ L6 Z$ Z0 f# d
get NULL1 long2 O3 H/ T9 _" a# B3 B8 Z
for i = 0 < FILES
3 r5 L3 o) g& l2 y, ]; P; {* Q8 Ugetdstring NAME 0x808 l8 d. A& f9 F6 M7 S9 t1 b
get UNK2 long
% a  J2 J' y0 [& W% h; ]7 nget UNK3 long, D$ p( p( {4 h7 ^( }+ H. o
get UNK4 long0 ^& S- k) N( b) D9 b& `
clog NAME OFFSET ZSIZE SIZE4 M3 C7 \, n6 W: H# S
next i
' S5 z" {1 T0 A' N' C
: ?5 p. s- I6 r% P# O
- l; v( ^9 ], o, Z; T1 jok this may look complex but it is almost identical to the first tutorial file except we added 1 more variable
- U& Q. R+ [! @: N8 YZSIZE this represents the compressed file size while SIZE represents the decompressed file size
# s: ]- P& B% A& w# gand we also changed the log command to clog to represent it is a compressed file.8 l* l" f2 J8 ?) e  C6 W! ?6 J2 R
: J1 M& o- m( O9 a
ok so now we have our loop and the commands to extract our files but we still need to fill in the variables. f" E6 O" P9 J" o
OFFSET ZSIZE SIZE
, C# C2 N* a& A( n3 [so that means our 3 unknown values must represent that but how do we know what order they are in?
% i( p7 d+ S6 C; s  t# Z( U: f; `" q4 R  }. g1 C) z/ P/ f
Well ill let you in on a cool trick follow the file loop to the end start at Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds and highlight the whole 0x80 length- G5 P8 h5 U! a2 O
then add our 3 unknown variables so that means we are highlighting 0x8C for our length. the first file is from 0x10 - 0x9B
; Y+ K0 K2 I2 I9 O0 k$ L. e- ^% vso now do this for the rest of the files and you end up at highlighting 0x1B4 - 0x23F: }( ?$ ^# j, c/ O9 L8 k9 A. w3 ~
  
! D- D$ a# ]6 c  eok so we reached the end of our loop now what?* f: I- I' Y: ]& i) h; n) q* P
well the next 2 bytes are 78 9C and this is an archive extractors best friend when you see this at the start of a file.& _" u! U$ N  [9 l7 r7 s1 P9 b; S8 f
78 9C represents the standard zlib compression header / V& n! }# q" `. I0 ~
ok so this means our first file starts there which is at offset 0x240) p0 m2 N! w$ @3 B" |$ \

* G/ x3 q  y4 }" n( i1 {$ \7 Gwell lets go back to our first file in the list and look at those unknown variables.
3 F, S* Q- I; |( A. ?, O24 72 00 00 is = 00 00 72 24 = 0x7224
8 W% S6 j( `8 F4 P80 00 02 00 is = 00 02 00 80 = 0x20080
5 U% O* h  y# A- `/ L1 V40 02 00 00 is = 00 00 02 40 = 0x240
1 W7 U; @# D! v+ nI think we have a winner so the third variable is 0x240 aka the offset
+ |3 l: o. k" \- O! u+ w& Eso lets update our script
: n) W1 b1 R9 X! Z3 D* ~
4 C8 M* A6 x# l, d) K4 k7 O
( `; d1 _/ @' s! L) A3 E# pCode:/ R6 z* j$ ^2 _  D' K3 S( Y# n
idstring " PSH"
' I8 M" u6 A: ~$ I' K# g9 E! ]9 Bget UNK1 long
2 }( q/ S% h" p* X4 Iget FILES long+ f: y2 e' b$ R! \' ~
get NULL1 long
4 @' R+ W7 g1 @$ m2 z( m4 Gfor i = 0 < FILES
3 ~( y) K. Q4 {getdstring NAME 0x80
; l) J3 V8 s5 f) Qget UNK2 long
/ @; q$ F# ]; N( lget UNK3 long
% [( N$ B$ q* |  ?" N" H: oget OFFSET long2 o' G* K7 i* t% n& [/ z2 o
clog NAME OFFSET ZSIZE SIZE
# Z1 k& x$ D' Z9 i, d* ]( k3 Z7 jnext i5 [. r* G! Z  ]. A9 x8 h3 ^( o

9 x. Q* E7 d0 @4 e6 {" I; l- O8 J8 t* w
now that just leaves ZSIZE and SIZE
) ]( S# w% B8 H: n% `+ Xwell bye process of elimination the decompressed file must be bigger than the compressed file so we compare the 2 variables
* B! E# J) v2 p! U0 B$ k5 B24 72 00 00 is = 00 00 72 24 = 0x7224
& h. {+ f4 z3 ~80 00 02 00 is = 00 02 00 80 = 0x20080
8 |5 Q% e. Z( h0 c- N  g2 @well 0x20080 is definitely bigger so we now know the last 2 variables& t9 v: ]5 b: ]: v% @8 S- W

5 y! ?$ U* x0 N7 ^/ m0 w+ X8 B  ]. h/ [
Code:2 o+ q$ m3 M, G5 k; O
idstring " PSH"5 d) T8 R# e8 d; ]: Z
get UNK1 long* T( O7 p) D/ ]' m4 e7 l" w
get FILES long( |; ^  \2 F- [
get NULL1 long
9 }- G* B' W1 dfor i = 0 < FILES
5 g+ F1 W/ M7 Q$ p6 X- M* c8 l( agetdstring NAME 0x80
; Z) a: c, u0 ~, O, mget ZSIZE long4 x+ k3 {: k1 a4 W9 A
get SIZE long# D0 `9 F6 w# P4 s9 i* g
get OFFSET long
, }5 C. d" _  ?& ^1 [  ]1 jclog NAME OFFSET ZSIZE SIZE
  ?0 D. s3 p/ z. S+ ]0 K0 b5 z- Z& |next i# o) Y1 h3 m% v0 n

0 K7 B/ Z4 x4 q& [$ [5 P
9 k6 }4 x4 Z) e5 d2 c8 fnow try our code out on the file7 X/ `/ {* _& N1 O
open the command prompt and change to the directory( {, s, E- |" r1 q
c:\temp' v6 |% D; T& p; b. d' \, ]* ]
now type
* c1 e: j" y$ X4 Equickbms.exe -l astro.bms BoneObject.hsp .  T, [4 d2 [. t) R/ i4 ^
yay it listed our files without any errors now lets try extracting them# b$ k7 {* Y" `; t* F% O
create a folder called extract
' w" s7 F, h& t9 m, q" Fand type7 _, P8 T# i9 ]7 w* f
quickbms.exe astro.bms BoneObject.hsp extract
3 e' V/ V& V/ X: l, x7 j- Z3 Eif we look in there we now have folders and in those folders are 4 pictures ( _: o3 n% x6 e3 P) ~5 D
we did it.
- s* r" u0 [5 D3 v1 ^, A: n" R5 F3 ^3 i% k3 F0 O. c' c
Let me know if you want more pictures or any way I can improve the tutorials.8 B3 ~: P6 M! }& I
: A' J9 d: f& x2 P8 n, q
3 C5 a0 t& l, V/ s9 E  _
Last edited by chrrox on Tue Jun 09, 2009 2:33 pm, edited 1 time in total.

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发表于 2011-5-29 19:15 | 只看该作者
要是大大能把QuickBMS的帮助翻译一下就好了[s:89]
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沙发
发表于 2009-9-9 21:53 | 只看该作者
cool turotials ,thx!
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板凳
发表于 2009-10-7 21:48 | 只看该作者
应群里朋友之请,翻译了一下这篇教程,不是完全照翻的,后面差不多是自己rewrite了,希望会有帮助0 C) F; Z  A9 \7 F
——————————————————————————————/ }& ~: l( L+ H& \0 r0 I
建立一个目录C:\\temp来放我们解包出的文件1 h2 M/ E/ n) |, P( R- o. V+ @

/ E/ i2 k. s' w+ T) Z1 y  |1,进入C:\\temp文件夹( M  r! T3 s6 H* |* q5 t8 c
2,建立一个新文件 astro.bms(QuickBMS解包脚本)
# y  r: \  `1 G! H/ B- h3,把最新版的quickbms也放到这个文件夹
2 _; w7 a7 B/ \# p" b1 Z8 p' h( ?7 K8 J2 z6 }6 H8 `4 _
现在,用你的十六进制编辑器打开BoneObject.hsp,来好好观察一下& `6 y9 r: A" f3 n
(图)/ i) I% \7 p2 Z
; Y5 Y  `7 x. A$ o& y$ B
很好,我们看到了一些清楚的文字$ d  |8 y) c# [, w$ [0 y- m- ]
- r/ i/ `; M! _+ W& f
你会注意到最开始4个字节20 50 53 48,是空格跟上PSH; o  g4 c, d- S: P
看起来就是文件后缀名的反向排列
: _: a. K7 r9 e& U& E* }1 Y0 u# G4 o. w! H
这被称为idstring(标识字串)
% y) v- p) L6 O2 |4 {7 J! U) B# v所以,现在在脚本里写上一句: z& Z9 b$ M3 ~

9 H4 N0 u% @: k% \8 k$ A$ W+ \get IDSTRING long       (将四个字节(long)存为IDSTRING)
/ G$ ^1 ~) {. [2 C( ^1 s. K
( b( R( B% E" i( N& Q1 V这没什么错误,不过我们有一条更好的指令% c, h2 m: I, h0 S5 f

9 @! \1 v) S+ l4 J' W0 K6 Widstring " PSH"        
" d# w/ W9 e7 `) }& K2 D' R9 I+ }; \3 F9 h" s- P3 r1 A" ~, M, S" j9 l. [, r
确保你没漏掉引号。
8 h+ X: ]+ P$ e0 c
2 \7 @: p+ C( R1 K4 w这条指令更好是因为你可以告诉程序,如果没有在开头找到这个标识符,那么就不要解包这个文件。1 a6 F8 M( y/ Q$ n# ?2 `

5 a! ~  |+ n9 c" E) b, @之后继续观察文件,我们可以看到% t4 u" ]8 p* o7 M
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds , Datas\\Texture\\BoneObject\\Toon.bmp , Datas\\Texture\\BoneObject\\Toon_a.bmp , Datas\\Texture\\BoneObject\\Toon_zero.bmp
: L  j1 P% Z5 I$ X# M所以我假设有4个文件在这个包里。9 P- L! R5 T% K1 G+ S2 {
2 ^, c; v# q8 e7 h
ok,回到开头标识符,接下来看之后的四个字节,是01 00 00 00,那等于00 00 00 01或者1,文件数量比这要多,所以我们不明白这代表什么* a8 Z- Z+ q+ w5 h
0 E+ M( \% H2 T& c5 d( M) @
那么我们在脚本里写这么一句- S' E8 ?# F: F( m, B, _# S
( m& L' w' _8 }, r0 F2 p! Z2 ?! R& `* C
get UNK1 long
- F  o, }* e6 F- ?% a: D这句指令把4个字节存为变量UNK10 Z# Z$ X5 M# \

- F6 c$ K: T$ Y  n* p0 Wok,之后四个字节是04 00 00 00,就是00 00 00 04或者4' f, a. i9 ~8 A7 E  v' R. Y& r
这就是包里的文件数量,所以我们在脚本里写这么一句:9 K( g8 M2 A: n# z: C

, f* _- D* ~; e/ U( Eget FILES long& Q6 e% x4 S; n! S; K0 P9 x
这一句把4个字节存为变量FILES+ ^: H5 u* ~, f3 P1 E+ W9 x' |

9 t! S1 p/ c. _: ]3 C之后四个字节是00 00 00 00,嗯,那就代表09 S# J# Z0 |) r: y  |
于是我们这么写
* n6 h- s4 h3 W$ p+ N
' V& x- H$ u" H" ?2 `% R" V& Sget NULL1 long
, W% l7 t8 D: p8 t% V$ B% `2 w9 a把这四个字节存为变量 NULL1, F) B8 f- Q7 B$ J9 `& W9 i
/ o/ ~- I# L7 C
好了,现在我们到达了第一个文件的文件名部分8 ^# i( R) d0 g& O" U  z
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds% M2 W9 H3 E/ [: @/ v
这个字串的长度是0x36,不过等等,这儿没有一个标示符告诉我们文件名的长度,那么我们该怎么写脚本呢?3 {' B3 L6 }1 {0 z" V

: z8 k- ]1 y' h3 B% r# ywell,我们来找找规律) ^# i( Q3 G0 z: y( C
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds is 0x36 5 J( Z" Q2 K* N' z3 |, C) o" d
Datas\\Texture\\BoneObject\\Toon.bmp is 0x21% v0 i) |6 f- P) h9 v* o
Datas\\Texture\\BoneObject\\Toon_a.bmp is 0x23
6 t* V, w9 r: f1 L# }+ j* UDatas\\Texture\\BoneObject\\Toon_zero.bmp is 0x266 P8 a% R: k' w% N; ]

! F& \2 y. k9 ~1 u3 o( x看起来没什么规律,呵呵8 W' ]9 y- y6 j% q% l

8 i* w4 _' e5 b4 b不过我注意到,文件名之后都跟着一大堆的00,那么把文件名加上那些0,长度是多少呢?7 W" Z7 r6 A. C1 h+ P8 \- v& [/ n

, M& x- k6 r) I) M/ O: i. Q% R8 S, b5 PDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds + 0's is 0x80   
8 W  o, q1 P% ^$ {3 h# \Datas\\Texture\\BoneObject\\Toon.bmp + 0's is 0x80 ! F' v8 X6 N2 j9 {' ^
Datas\\Texture\\BoneObject\\Toon_a.bmp + 0's is 0x80
6 P# U& b( T+ mDatas\\Texture\\BoneObject\\Toon_zero.bmp + 0's is 0x80
# U, Y+ Q1 m9 `( ~
# M; L9 T! q, t. ?# C/ S: r4 }嘿,看到了吧,他们的长度都是0x80
* P/ _( g: P* `# a4 F. _所以,我们在脚本里这么写! X! K/ E. ?5 A9 A8 t
getdstring NAME 0x805 [) d5 N' p7 p5 \& D

- F+ j) A/ @2 ?0 X% N5 r这告诉程序,读取0x80个字节,把他们存到NAME变量里,程序会自动移除后面的那些0
8 y) Y8 g; X. h1 V. ~- f! s- e7 b* W/ z7 D# \# e8 m9 F8 c. P# g
ok,那么在下一个文件名之前,我们还有0xC字节的数据,这些是三个long型数据
' y9 W9 ?3 ?! I: z* k. \% |+ r! ]我们暂时这么写,之后再来搞清楚他们到底是什么意思
: A. Y, U3 e9 l6 E% e6 `% nget UNK2 long7 [6 I* e1 e, b2 S1 L
get UNK3 long1 A! W6 V$ Q, T1 ?" _& V3 a; n- U+ O
get UNK4 long* ?3 h1 u% g" R2 C- l8 P& K) b4 ]. ]8 J
# M2 @' G9 x6 b7 `  @8 O
那么我们现在又看到了文件名
2 p4 ~' x+ t) E! V1 [7 O7 c现在,我们找到了规律,所以按我们之前学到了来写脚本:
8 s1 S- `3 N3 E1 p6 U' v. C
' G* W( z. ~  x代码:" u! }0 N' w: R" {" ~
idstring " PSH"0 t' C$ I" V. y+ ]$ [; D
get UNK1 long
5 g, W9 r. s* }$ m" n6 n0 x/ Zget FILES long& M9 u# L" F, q+ B* j6 l5 t
get NULL1 long; H0 P9 J+ |2 \. o. e  Q8 o- y
for i = 0 < FILES- g& `+ r8 D: g6 z+ S
getdstring NAME 0x804 M; v& U5 k: F$ ?( _# F
get UNK2 long' r+ G, h( @2 @0 N
get UNK3 long : K7 Y8 O( |) G( z
get UNK4 long
( _! _/ |: z: F8 U$ F; Eclog NAME OFFSET ZSIZE SIZE- o7 g% j! L+ o; p2 o& t9 V' g
next i
/ a: y& g/ @- @; H' w  l* r
  P0 K% l8 a: `: }: Sok,这看起来可能有一点复杂,不过应该跟第一篇教程差不多,除了我们多加了一个变量ZSIZE,它表示压缩过的文件大小,而SIZE代表没压缩过的文件大小& s2 k, I% L0 ]0 `# L
我们同样将log命令改为clog,表示这是一个压缩过的文件。# |6 \9 }' C+ x5 S' H

( t. O; ]% ]5 l  e  ?- \现在,我们有了循环,指令来解包,不过先得给这三个变量赋值
# L% v! @+ y$ V. U. EOFFSET ZSIZE SIZE
5 R: S  d9 @. E" i; C8 ]- }) j# V0 |& u$ a' e3 T; Z
这意味着我们那三个未知变量很有可能代表的就是他们,那么我们怎么知道顺序呢?" s; X+ {  o( H

1 ~, ~/ f# q" v: Q) Q; n) z好,现在让我们来到这个循环的末尾,定位到最后一个文件的文件名,选择0x8C个字节。
, c. u1 A+ v0 G+ _( F  {# p& r$ T然后之后2个字节是78 9C,这是一个解包器的最好朋友,尤其当你在一个文件的开头看到它时。/ f! E, |% {2 e
78 9C 是标准zlib压缩格式的头部标识  U' \8 d: b. A5 @
. c! l, _6 L7 J! n
所以,这意味着我们的第一个文件从偏移0x240开始7 J$ F  }' q2 c9 T
- {/ H  ]- n/ g. N
之后,我们回到列表里的第一个文件,看看这些未知变量。
( S  W( c3 @% O2 d) T8 b24 72 00 00 代表0x72246 i# H- a- I0 Q- ~7 f# K/ k' f
80 00 02 00 代表0x20080
' W/ e* H; g3 T6 w40 02 00 00 代表0x2408 ^' N; ]& b* V; P
8 t6 x. T+ n) ~( h& x& g
我想我们至少知道第三个变量0x240代表着偏移量- k$ X6 F  U4 E$ T( o
那么,更新一下脚本:
7 V) U( i6 B6 p" o% H% Y; g( C3 O$ L3 T
代码:   O& j) k# J  H( B3 m9 h
idstring " PSH"4 J5 V5 T& O- d& c* M; D
get UNK1 long
/ J7 h# G7 e. J% E8 S8 g3 Gget FILES long
6 `3 K6 P9 m; Tget NULL1 long
3 q$ ]* w0 I7 U7 c8 V3 a7 D9 lfor i = 0 < FILES
; Z# b! {7 N. D2 _8 wgetdstring NAME 0x809 a9 ?, m3 |5 n- O
get UNK2 long
+ O! m' d% x6 l# _5 nget UNK3 long
' [% L5 \9 A+ c3 [' gget OFFSET long2 M8 K: ]$ d0 h/ a
clog NAME OFFSET ZSIZE SIZE + X( s8 e1 s& d( Q; w
next i$ h- }: G/ Y" D5 Y
9 q' ?$ e  y' z: I% ]* X: [( Q  y
好了,现在还剩下ZSIZE和SIZE
: r& K/ b6 o. i  B9 N4 _' h2 h显然,压缩后的文件大小要比原来的要小,那么比较一下这两个变量
  e+ m" ?" x+ z  A: `一个是0x7224,一个是0x20080
3 b7 z2 K$ b3 m1 j' F% E. g显然,后者要大,于是我们这么改写脚本:1 Z8 z# ^7 w. j8 O- q' K' ]
0 M& t6 H( i) ~4 ^  U; ~" t2 j# l
代码:
% ?6 H: ?" i8 {, A6 Widstring " PSH"4 R5 ]* I* O1 N% `1 M* V
get UNK1 long
' w( u+ Q/ B) D' I/ P% nget FILES long  j& t+ C6 M8 t4 H% r6 x
get NULL1 long- X% b5 l. C8 B6 p7 |
for i = 0 < FILES4 P, f0 C, I. f: ~3 a: q
getdstring NAME 0x80. @5 Y8 R' m' p2 `! z) A
get ZSIZE long, R( W5 ^8 x8 |: G% {5 g# ]
get SIZE long$ `  L: w2 K7 [; [0 W& W
get OFFSET long
2 Q: {+ a, D3 ~. K) x5 \clog NAME OFFSET ZSIZE SIZE
$ M/ Y$ }! q& _' @7 C) b4 nnext i- Z, C& k$ d4 `4 }
7 i! ]% X( d5 w* P: Z7 C5 }6 W
好了,现在试试我们的代码吧。
" _2 }0 ?( C' `$ Q/ v  [' ^4 e打开命令提示符,进入到c:\\temp目录
6 F8 k$ s) I" ^: q+ m6 Y* C; h( u0 l5 c3 {. O8 ~; v
输入 quickbms.exe -l astro.bms BoneObject.hsp
+ C3 i/ e6 R% E& Z8 D* a; t4 `) R- W他会列出我们的文件,没有提示任何错误
4 h$ Y8 P. P* k$ F2 k  l- z( O: c好了,现在我们建立一个目录 extract
- Z+ p1 D8 B# Y' B7 }+ x- g3 {输入quickbms.exe astro.bms BoneObject.hsp extract3 e4 y: h; O7 A& @+ k. _7 I
4 s7 P' G3 Y$ s+ E6 A3 S
好了,现在我们在目录里有了4张图片,我们完成了。
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笨蛋狐狸 该用户已被删除
地板
发表于 2011-4-28 09:42 | 只看该作者
感谢各位前辈,留爪备查。
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