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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
3 E* H7 _/ W7 A) H8 E" v! I1. to c:\temp1 ~+ x0 p2 Z, h/ j8 a  `
2. create a new text document called astro.bms
; ^+ u" X! F7 [3. and place the newest version of quickbms in the folder also.4 E+ I0 }1 }$ L0 [- L5 i

7 F& |$ d% v0 GOk so open up BoneObject.hsp in your hex editor and lets take a look at it.
$ e3 T- c! N+ s! i  
2 m: `5 B8 E" U) Ogood we have some plain text.
" u' P6 q* u7 J$ A% Byou will notice I highlighted the first 4 bytes 20 50 53 48 or " PSH" that is a space followed bye P S H.
7 A) K# _7 N8 r, {5 Z% C0 Y6 O$ U, yhmm that seems familiar that is the file extension only backwards. this is know as the idstring0 r) i/ I7 f+ T: c7 u: V
so up until now you would think to write in bms
" _, Y5 Q" q% }9 I( M+ Dget IDSTRING long
3 h: V" D1 w% d5 ?9 [there is nothing wrong with that but there is a better command
' e" c; W7 `! v- X0 l1 ^& cidstring " PSH"
. O2 f/ G8 |) I! G1 t0 n4 B# V' pmake sure you include the quotes.
0 B1 V# N1 ?! hso open your bms string and on the first line type/ _5 z" I* N$ @: M
idstring " PSH"6 V8 b- J2 h! U5 z( z, H
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.3 L; y7 ~8 Y( ?; a8 q3 Y. g4 s; E
"aka noob proofing it"
+ \' ]/ g2 h! R1 i% U' n# b
: P7 V9 R* O0 z/ Z$ m! k' LOk so now lets look at what we can read I see * f4 U" R$ n; Z6 [9 U
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- ]: q2 k0 u+ c6 a
so I will assume there are 4 files in this archive.* W9 |3 p8 s8 D  C/ ?
well 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& L, c7 v! c. E( R+ ^, V: D
so lets write that in bms language
* @4 Q" k8 _$ Y5 V/ nget UNK1 long4 i. u1 _6 i, J8 V' f# A! ^# @
this saves those 4 bytes as the variable UNK1.7 D$ u$ q8 V+ q& `: q7 v$ I

4 f0 k% ]8 y$ m) ?1 p6 iok the next 4 bytes are 04 00 00 00 hmm this translates into 00 00 00 04 or 4! Z& Y9 @- o8 u1 F% E3 Y. X
hey that is the number of files we counted so lets write that in bms9 m( z# O) K" o( `. w
get FILES long
/ [; C& _% @1 V- _) Kthis saves those 4 bytes as the variable FILES.( c3 y7 J% Y- G9 S# b2 K6 @' I. b
: a( @; R  x7 i, [; T( f" o8 g1 T1 D% B
the next 4 bytes are 00 00 00 00 well that is equal to zero so for now I will write that in bms. `8 K! c6 f3 N$ T: r) l/ C
get NULL1 long5 q& N9 K6 {( T; j: Q8 d
this saves those 4 bytes as the variable NULL1' Q7 H+ r# Y0 J, l0 o8 b

; q! m- I0 i9 aok now we have reached the first file name Datas\Texture\BoneObject\npc_nagoya_octopus01_body.dds/ B0 L# n3 _5 j
this 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?
0 J' e% Y; b4 e; T1 \well lets look for a pattern
+ a# _& g$ c, j; Q8 QDatas\Texture\BoneObject\npc_nagoya_octopus01_body.dds is 0x36
) Z: R# e1 i/ i! R, j. e: [Datas\Texture\BoneObject\Toon.bmp is 0x21
0 Q4 m; H$ a" O2 gDatas\Texture\BoneObject\Toon_a.bmp is 0x239 S. c' A; t' _+ F0 j2 a8 k
Datas\Texture\BoneObject\Toon_zero.bmp is 0x26
4 ^, x% G- t0 S8 g2 V: \1 a6 xhmm I don't see anything that makes that a pattern.
- B/ v7 J, Q+ D: dbut I do see all the names are followed bye a lot of zero's. how long is the name + the zeros of each file?
9 G# R6 I9 q; Q, d3 E; GDatas\Texture\BoneObject\npc_nagoya_octopus01_body.dds + 0's is 0x80
* H5 J5 r, M; a% E( l( KDatas\Texture\BoneObject\Toon.bmp + 0's is 0x80
& Z& N( `$ W: ~2 V! eDatas\Texture\BoneObject\Toon_a.bmp + 0's is 0x80
7 u+ c; w) X3 j/ i( Q+ qDatas\Texture\BoneObject\Toon_zero.bmp + 0's is 0x80
' W, c" z5 m0 Bhey they are all the same size when I include the 0's
  u: ]/ a3 {% i& o- Fso in bms I would write this as
# a! T  _. p4 L3 I" i8 Hgetdstring NAME 0x80: D' _, b; @# @0 Y" P
this tells it to grab 0x80 bytes and store the text value of it
7 Q4 C, a( G# f# }; Y  q' Nand as an added feature it will automatically remove trailing 0's
( `% o' e4 {6 t5 a; y+ v0 _& e+ q& M$ |
ok so now we have 0xC bytes before I see the next file name
& K% X1 N- c; P8 ?6 N# |; |which is 3 long values
3 a- j# U: ^" e. [so lets write those in and we will figure out what they represent later./ Z! c1 |. `( N4 ^
get UNK2 long# u# }5 B; n2 Q0 N5 w: r
get UNK3 long
3 G, |- j8 V& d' ?% G, j5 Mget UNK4 long
9 U- {$ D* Z* L/ I' c) X
" a$ u) r/ o( `1 `1 F. Z* b; l  Dok so now we see the name again
9 R) h. B! |- q( x$ F' xwe have our pattern so lets write our script based on what we learned, j! n, B% B: S  ~
so it would look like this up until now( u1 @5 `! c3 I, `9 f+ `

7 T3 `; I9 D4 q+ ~0 sCode:
5 a+ w: \& C$ C6 d+ M7 q! w5 Yidstring " PSH"
8 z8 l. u5 v  b# L9 L3 u& |% {get UNK1 long
$ m7 b2 |7 F( V) d; n# Sget FILES long9 v8 }: v) h& Q* y0 g1 I7 }
get NULL1 long8 C9 g* p7 R3 G7 U' Y; V3 V+ @3 d
for i = 0 < FILES
! p+ e: j' d- w* F2 h2 b* I2 Ggetdstring NAME 0x80( e: m( f7 w( L& B
get UNK2 long
8 n8 Q% K3 f# _- |get UNK3 long+ }) g: v8 m0 G3 Y5 a
get UNK4 long
$ g" l4 |$ A& R; {clog NAME OFFSET ZSIZE SIZE
! s- t; o+ S2 b0 Onext i
' H0 L7 e5 [, H/ Q" \0 j
6 P( i4 w6 K" I6 D" P3 C& g4 I, F. L) K5 O9 O; T+ N, ]) [$ U7 ~( Y, e
ok this may look complex but it is almost identical to the first tutorial file except we added 1 more variable
: Q% d/ t/ F5 v$ f6 w- H/ N' xZSIZE this represents the compressed file size while SIZE represents the decompressed file size1 W6 U! m' u- E' I
and we also changed the log command to clog to represent it is a compressed file.- S* {8 a2 _" d

+ @+ o7 p2 g# Q' Z: ?! d9 E0 R. M$ Fok so now we have our loop and the commands to extract our files but we still need to fill in the variables* Z9 e6 Q' c1 V! [/ \& p
OFFSET ZSIZE SIZE# m" H. S* z! d2 f0 \; `) }
so that means our 3 unknown values must represent that but how do we know what order they are in?
4 p/ Q5 y* D; a1 l2 V5 z
6 K& s; ~- J) g# v# s, z: ]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
+ |4 L9 t( y$ {5 I- V9 Fthen add our 3 unknown variables so that means we are highlighting 0x8C for our length. the first file is from 0x10 - 0x9B
) n, W4 m4 Q4 N8 @so now do this for the rest of the files and you end up at highlighting 0x1B4 - 0x23F
& _" c0 G) z4 S& r- P# E- @& p  # t+ }7 c  w) O7 H. R
ok so we reached the end of our loop now what?
, ]) }7 _! X* v5 r. cwell 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.
/ _! ]  P3 l7 k  H3 j+ X78 9C represents the standard zlib compression header
+ ^, l1 N! i( D0 L4 |5 kok so this means our first file starts there which is at offset 0x240
- ^& }7 d( m# T* \# k! r5 B
- y" ?9 ?3 ]# P% d! N; m# Zwell lets go back to our first file in the list and look at those unknown variables.
0 h5 X7 `) h3 u/ r24 72 00 00 is = 00 00 72 24 = 0x7224& o9 K; D/ e4 |  y7 K; N- |# \7 W
80 00 02 00 is = 00 02 00 80 = 0x200809 c2 Z, L8 T, x  K2 T; B# @
40 02 00 00 is = 00 00 02 40 = 0x240
' T! a9 b! y$ b! g" A. MI think we have a winner so the third variable is 0x240 aka the offset8 m4 S, ~: k1 z9 R
so lets update our script
( R/ p" i. A. _* U4 ?( c% b$ V& S9 n
- R  e- a) s8 w2 L) e0 B6 f: m' F( H+ U" W
Code:& K2 f& ?0 s7 w. v: w  p$ a
idstring " PSH"7 O9 s) T* {' Y$ v3 B1 z
get UNK1 long
( G. J1 w/ T& s% G/ [get FILES long7 W1 K1 m' r! e6 N/ R( U' }  ^. Q
get NULL1 long. `2 C# w+ U/ g, i) M0 W
for i = 0 < FILES6 {- @& N; S% z8 V& S: K
getdstring NAME 0x803 ]3 i& s  P6 ?
get UNK2 long
4 h  ^; |0 `% d, Vget UNK3 long
' A/ N5 k) {' v; o6 Mget OFFSET long
/ J5 q% c( r: P$ p3 \; tclog NAME OFFSET ZSIZE SIZE' M3 m8 y; C& ~) S1 |- h
next i
' U- i& j( n! H. G; F5 i% c* w) m& U* i3 R' X% i
" o6 x0 _& l* V: P
now that just leaves ZSIZE and SIZE
1 l6 l  R6 Z; w+ `: G+ D' cwell bye process of elimination the decompressed file must be bigger than the compressed file so we compare the 2 variables* X8 T2 V/ m5 s0 L
24 72 00 00 is = 00 00 72 24 = 0x7224
3 p7 l; v* E5 J8 J8 \80 00 02 00 is = 00 02 00 80 = 0x20080
; Q# a/ L9 I% ?) z2 Vwell 0x20080 is definitely bigger so we now know the last 2 variables6 u' a$ G$ N* A% a) w

2 P- Y, h, V5 d1 S4 c" f' O
: f0 u; H. ?  X# p, }Code:
1 n8 q1 G* B5 g0 m; X7 C5 N3 tidstring " PSH"
1 e7 ~8 [2 c3 {; ]4 p! `4 {7 Kget UNK1 long
/ S4 \( b+ Z$ i& V. Y' vget FILES long2 T7 f& t7 j2 `4 l
get NULL1 long) _# [& n" A& Y& ^
for i = 0 < FILES
$ H( ~$ b0 T7 Xgetdstring NAME 0x80
5 M2 h; G, Q" |: U3 eget ZSIZE long3 D  r- k# Y6 m# Y( N8 j4 G& ?
get SIZE long/ H% s: T) G7 J) I
get OFFSET long% p4 y9 e/ n9 y3 B1 v! }
clog NAME OFFSET ZSIZE SIZE. N9 n$ G/ U$ [- J) ~/ Z) |
next i( L8 ?- y) s# j# {0 x( S+ Z+ W
) `. u, w8 R/ ]+ K
$ B9 G- u+ L& f# |2 v
now try our code out on the file# a9 s& @  `' j; H2 j# w9 N
open the command prompt and change to the directory. Q7 K3 X& d: |& c: v. y
c:\temp( @' a% f0 S7 O
now type
$ O! ?* s8 r, N: ?- x4 \quickbms.exe -l astro.bms BoneObject.hsp .
* w# }3 C+ p2 k& [' uyay it listed our files without any errors now lets try extracting them
  U+ }- x) S+ v. f. C( p# {create a folder called extract6 J) w( g  [3 Z) G+ X$ {2 J" v
and type  O! q9 B! _% G1 X& |) h- u7 ?
quickbms.exe astro.bms BoneObject.hsp extract5 F/ Z6 y$ q' Y" o+ x% i9 r2 \1 i
if we look in there we now have folders and in those folders are 4 pictures
3 C9 h" q9 l# }  k+ I" H' q2 Cwe did it.
  r: S) \8 x9 {' Q' Z0 a: {- _- B/ l3 m
Let me know if you want more pictures or any way I can improve the tutorials.3 n# C7 X9 a9 A

: S4 N# T6 a- e) Q
' y$ @. f8 P6 @; n+ {( B+ hLast 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
' y" c' `5 z' B4 G; y6 C——————————————————————————————5 I/ r. w. Y" \- T! X1 ]
建立一个目录C:\\temp来放我们解包出的文件+ U: v$ P) x% m% H/ K

: H# H% c* V/ K) a1 B7 g1 U% a1,进入C:\\temp文件夹4 o; e; b3 S2 B) h9 P1 c
2,建立一个新文件 astro.bms(QuickBMS解包脚本)1 z" Z) g3 {7 S
3,把最新版的quickbms也放到这个文件夹
$ a8 i1 G, _6 C3 R4 K* c- W
$ q' r! m, K+ Z, _: m9 I现在,用你的十六进制编辑器打开BoneObject.hsp,来好好观察一下
( U6 x3 d9 r2 E9 ]$ a# A) g3 k(图)! i  p' u& n% Q

* D- g/ z# }" A# s; i8 i8 Y& g很好,我们看到了一些清楚的文字
- {: A  p* V8 r2 ?# V. v0 _, P* K: g, b/ E  X. c
你会注意到最开始4个字节20 50 53 48,是空格跟上PSH7 j* l3 g* v+ G% ^/ ]* `( q+ l
看起来就是文件后缀名的反向排列
2 t5 m* [8 o  T( d
6 {  q: @" [7 x, E" H3 y这被称为idstring(标识字串)
* T7 j+ d: J" [: T3 R! k. g+ N所以,现在在脚本里写上一句+ B4 @+ f# E( Y) t" A4 Z/ X3 U  w; o
& n+ C6 r# ]3 l% A
get IDSTRING long       (将四个字节(long)存为IDSTRING)/ M' n0 i1 h6 v2 Z) D& d; J  N

/ Z  F5 O# |1 h7 s3 O* s$ y这没什么错误,不过我们有一条更好的指令
3 w* {+ k+ S+ I) w! _9 N( m" O/ L% B8 D* K- a
idstring " PSH"        2 a% O, Z/ B, b  N0 }1 b
; N, u$ ^3 Y8 q' I
确保你没漏掉引号。5 P) t. Q2 Y- v. G( t
9 o- Q( n, a+ }
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。
5 i2 `. G5 ^  e6 x1 j8 y) `8 Y4 F5 P" E8 y  H1 n+ c
之后继续观察文件,我们可以看到
+ {( l/ C6 W0 ^" YDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds , Datas\\Texture\\BoneObject\\Toon.bmp , Datas\\Texture\\BoneObject\\Toon_a.bmp , Datas\\Texture\\BoneObject\\Toon_zero.bmp 2 W  k: r  @3 a  V+ m
所以我假设有4个文件在这个包里。! Q; F6 ^! W- M, R" s  x

2 P1 }/ I( H: l4 t- aok,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. Q7 g. E' L3 A- D0 C. B
+ e, Z; B; J" Y% a! \! i. c- Z* z
那么我们在脚本里写这么一句) H4 W& D: l8 S. {, x7 j+ V7 c
+ P) H5 I9 t% k) n4 Q  S  ^
get UNK1 long
* w8 P3 s7 P0 x' v这句指令把4个字节存为变量UNK1+ _# ~) B& u, i/ e' D  W* O

$ c  Z9 F$ H  m' r' A! G0 t1 R( hok,之后四个字节是04 00 00 00,就是00 00 00 04或者4" ?2 M5 I- T( `9 O/ M. ~' Y
这就是包里的文件数量,所以我们在脚本里写这么一句:
( i9 L# ?4 K6 e: Q, p) e) V
* n6 L, O! K2 m, Jget FILES long2 O- m# ^9 n/ W! C' p  b" M
这一句把4个字节存为变量FILES
8 G2 }7 |2 F$ M7 N  e1 _; i/ y) B1 ~' Q* p% i
之后四个字节是00 00 00 00,嗯,那就代表0
2 `* s% Y: @" @) ^. F于是我们这么写. I- J4 _) g  z) _
2 B2 z) |1 H6 y% D9 J( {( P. _
get NULL1 long
) [% t$ o3 y5 q# q- V把这四个字节存为变量 NULL1
+ M5 y! w0 \* p- k8 t# ~% \, U8 w% C4 D( S& y
好了,现在我们到达了第一个文件的文件名部分, g6 G5 y) {' y" p& b7 l! I2 K- V6 j
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds3 R  w9 U* Y6 c' W6 W6 B
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?6 A, Z  G4 f* d0 Z7 g
2 \" V3 Y- l$ Z8 _4 ?( }" b' L! m
well,我们来找找规律# ]1 m8 P! Y, U6 U( P
Datas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds is 0x36
1 C7 m; v, x4 p: d+ R6 }& c! [Datas\\Texture\\BoneObject\\Toon.bmp is 0x21
! U8 \5 S/ k* X: kDatas\\Texture\\BoneObject\\Toon_a.bmp is 0x235 y5 ~. ]1 y# M0 c- F
Datas\\Texture\\BoneObject\\Toon_zero.bmp is 0x26
: W+ @0 a: _  u$ V2 n* o2 Z8 }3 y1 }' l( u9 i9 w
看起来没什么规律,呵呵' c2 T& ~0 `, W4 g2 Z) E. R" H
8 n. D/ `  ]* |( J( L: x7 @! S
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?
% C/ B9 }9 I$ {1 R/ v5 _8 P
, |5 P4 o: Y8 f+ `1 lDatas\\Texture\\BoneObject\\npc_nagoya_octopus01_body.dds + 0's is 0x80   
' W' v. y5 q( T% O! f4 L; M: W2 EDatas\\Texture\\BoneObject\\Toon.bmp + 0's is 0x80 / s- m5 ~# L4 ]9 f0 L- i
Datas\\Texture\\BoneObject\\Toon_a.bmp + 0's is 0x80
2 f3 O5 T- \/ O( bDatas\\Texture\\BoneObject\\Toon_zero.bmp + 0's is 0x80' l8 c; t! \) {: c

: h! Z4 |* K  m4 [7 Z4 {" I5 L* D嘿,看到了吧,他们的长度都是0x80
* w. q7 U; X! O/ [# F, L' a) [所以,我们在脚本里这么写$ i4 R9 ^) K4 }8 I
getdstring NAME 0x80& }5 |& e' l/ y1 Y4 a

) V$ [0 d/ G2 ]  g, y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; P) k2 h& D( K* v' M+ N
  l" Y/ Y# Q& F- f3 c( V. R
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
* g) x( Q$ W6 H( \/ R; N0 ~# z0 S我们暂时这么写,之后再来搞清楚他们到底是什么意思
  t! C) s% x5 Z( Gget UNK2 long0 h- Z" h. o! z/ Z! _
get UNK3 long9 E& P1 v. e% ?  @, C
get UNK4 long
: N+ s2 @# i) q2 [  ]0 x$ [2 G9 B. Y3 z3 j5 J
那么我们现在又看到了文件名8 Q- c1 H  Z2 w, u* \8 ~& e
现在,我们找到了规律,所以按我们之前学到了来写脚本:% D3 |5 Y! F1 ?. f: y

* @7 M4 t4 Y3 f; ^9 M. U5 [代码:
. r+ |* v) p& L0 D" |idstring " PSH"9 @* y5 z* ]& x& Y. H0 U7 p" x
get UNK1 long
+ B: ^+ Z. X6 gget FILES long
8 Z" o% Q; f; a8 E( P9 g$ \% ~' gget NULL1 long% C! U7 R; N9 X
for i = 0 < FILES# ^- C5 j' ~# B. I5 s6 H/ P; N, F
getdstring NAME 0x805 K1 o6 i0 K) S7 [' @
get UNK2 long; `) j  P* Z' N7 l  X
get UNK3 long
9 A) h# J- h* |) a1 q2 fget UNK4 long) L% _+ _8 `: U5 G6 x: y& v
clog NAME OFFSET ZSIZE SIZE
, A( P' ^5 [# {0 w! v9 p; j+ Rnext i" o) m$ K7 S4 y4 l# B4 T6 m' _

, P3 f3 Q: _- w4 H- Aok,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
7 d) F2 Q7 ]" r6 ?我们同样将log命令改为clog,表示这是一个压缩过的文件。
% m; A6 ~  t$ `4 t: a% P4 j4 C- ^7 j2 M- c, |+ |. n
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% D% O; v' f* T9 b4 L9 P
OFFSET ZSIZE SIZE
. |; Z3 X& @1 u6 }4 S* H9 m6 [1 H; A# X3 {7 t2 K1 M2 B9 P" X$ M4 ^
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?
! L/ b: j  W( `! x, Q0 e
4 C/ t* f  }* `' 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。. r+ N7 Q. z% [. F* l, O
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。
8 b. s# W% i, v1 \7 l78 9C 是标准zlib压缩格式的头部标识
4 A* v* E7 o+ E% R
6 M5 h( {+ f' n" m所以,这意味着我们的第一个文件从偏移0x240开始4 c* {7 B6 m3 H, z. V
9 l0 E* v& n( @  T. @5 v. r# Y- z
之后,我们回到列表里的第一个文件,看看这些未知变量。
: ?! }1 j/ G% \, q9 @6 {* a24 72 00 00 代表0x72248 E3 f; c3 R; Y2 J" i. `$ G) z
80 00 02 00 代表0x200801 H7 t8 v: _. Y8 O0 J
40 02 00 00 代表0x240
4 z# i) j" R" Q9 E
7 K$ d  Y- H5 f我想我们至少知道第三个变量0x240代表着偏移量& `: e0 M) ^: ]6 p4 e# U
那么,更新一下脚本:4 l4 ?; d' x/ V; H1 @( Y( I

( I; ^' ~7 }; r代码:
5 \/ p6 ^* p3 A$ ~8 |idstring " PSH"  r+ x, j! B& d: {! o9 j. P5 {3 Y
get UNK1 long
- {. B& {) m7 nget FILES long
/ L0 J7 ~# Z* Q3 yget NULL1 long' T0 I; d; B. R4 T
for i = 0 < FILES( n+ Z4 J1 a# t/ b
getdstring NAME 0x80$ R- B) }: F! k! d
get UNK2 long' F" X& Z" Z, k* V
get UNK3 long: P' I3 Z) \1 M& \" r
get OFFSET long' t( i  [6 v2 t
clog NAME OFFSET ZSIZE SIZE 5 v5 F. K  d, o' f6 p7 U. k/ j
next i
/ I7 Y0 ]4 }. {  B1 J2 `& r( N- m" j: \
好了,现在还剩下ZSIZE和SIZE( c+ A) ^- u; X9 ]+ u
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
: L5 h8 G' r  A" p* L一个是0x7224,一个是0x200801 ~2 d1 [! H- V& @( }
显然,后者要大,于是我们这么改写脚本:1 Q% h* }  H4 @2 I# m3 o) r

: u6 G& m3 B8 h# b  |; t代码:
3 g( B; v- A' d! d7 ]( ?& ?idstring " PSH"! t8 b- `2 f, Z2 @/ ^( e" w. _
get UNK1 long
4 r* A  |+ T0 i1 U( \. Aget FILES long
6 c- ^0 [- i) e9 P: L) rget NULL1 long/ `% X9 ]1 V2 i6 H; F# b
for i = 0 < FILES$ w( O: s- d, N! K( p+ R# f
getdstring NAME 0x80
; b+ `$ Y& Z. h; Rget ZSIZE long
+ [0 r: T; S! ]# k4 `get SIZE long
2 L) N3 @/ I5 |! X: Uget OFFSET long5 {& c* O' r" z3 T) a) o( p! U7 D
clog NAME OFFSET ZSIZE SIZE ' [9 U: j" W6 q; q( Z9 P4 s5 A0 \9 R! k
next i
) Y  U4 f* W; z) Q& ?  a3 E4 m1 Z+ i4 W& K) U! r4 ~
好了,现在试试我们的代码吧。
' G' J  @; x0 z打开命令提示符,进入到c:\\temp目录
: h+ E6 P7 H2 P! u2 `
& d1 ~  ~" V0 h1 D输入 quickbms.exe -l astro.bms BoneObject.hsp
; l. C" S) m% G/ |3 c! V$ a他会列出我们的文件,没有提示任何错误0 |# G  O! V1 A
好了,现在我们建立一个目录 extract
) n  @' a4 C8 L" Q" [; o输入quickbms.exe astro.bms BoneObject.hsp extract8 R; K/ }* N7 R2 X7 g

  T8 v" P+ S# S  E# T3 y. K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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