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使命召唤11新任务加载到一半弹出unknow block compression type:0
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为了帮助网友解决“使命召唤11新任务加载到一半弹出unkn”相关的问题，中国学网通过互联网对“使命召唤11新任务加载到一半弹出unkn”相关的解决方案进行了整理,用户详细问题包括:游戏是从网吧考的，一关都玩不了···，具体解决方案如下：解决方案1：unknow block compression type是未知压缩文件的意思你考的时候先把游戏压缩，回家再解压建议你还是到网上下载解决方案2：我试一下。下载太耗时了。。。通过对数据库的索引,我们还为您准备了：答：补丁都打好了吗？都打一遍试试。===========================================问：使命召唤11：高级战争，新任务加载到一半就卡住不动。。求解决办法啊！...答：网上使命召唤的C组补丁去下一个 然后 游戏运行的时候360什么的都关掉 360有时候会把游戏的有些进程当病毒 关掉===========================================问：使命召唤11：高级战争，新任务加载到一半就卡住不动。。求解决办法啊！...答：这个是游戏文件安装不完整造成的，重新安装下试试，提示的是读取不出来===========================================问：使命召唤11：高级战争，新任务加载到一半就卡住不动。。求解决办法啊！...答：游戏环境未安装，要在游民下游戏环境安装包===========================================问：选择完关卡进入的时候播放动画会加载一小点，然后进度条就会停住不动，...答：网上使命召唤的C组补丁去下一个 然后 游戏运行的时候360什么的都关掉 360有时候会把游戏的有些进程当病毒 关掉===========================================问：选择完关卡进入的时候播放动画会加载一小点，然后进度条就会停住不动，...答：游戏里面的任务前的动漫位置出现错误 建议从游民星空下载玩===========================================问：选择完关卡进入的时候播放动画会加载一小点，然后进度条就会停住不动，...答：照提示是叫你重新装DirectX11 好像损坏了（上面说不可恢复）===========================================问：选择完关卡进入的时候播放动画会加载一小点，然后进度条就会停住不动，...答：下载电脑管家，来解决！！在 工具箱的电脑诊所里 ！！===========================================问：每次进游戏读条的的时候都会弹出disc read error imagefile48.pak对话框...答：估计是安装出错，重装一次试试看，如果还不行就只能重下一个安装程序了===========================================两个问题: 1.你的使命召唤4不完整,缺斤少两,所以不行,我看好像是声音文件出错了!你去搜狗搜使命召唤4找到1G联网补丁,打一下再试试! 2.1.4版本?能玩吗?你在哪个地方啊?...===========================================也就是说你这次玩是强行关闭的游戏对吧,那你再打开游戏,他会提醒你弄到安全模式,你再试下。 。。。。。。。。。。what fuck!。。。。。这个。。。。。你确定是卡死不是...===========================================是?我新兵刚通关· 这是全部进度,包括任务完成程度、其他挑战模式和反对楼下的,挑战在剧情左边专门有一栏。 不玩不就得了……=========================================== 重新安装一遍。。。或者重新下一遍。=========================================== 建议你去游戏官网的下载界面最下面下载驱动安装,看是否能解决这个问题。要么就是你系统的问题了,你也可以重做系统。=========================================== 哦,,我和你的状况一样,下了个通关档覆盖了自己的档,然后选取刚才自己玩的呢一关,继续玩就行了===========================================楼主装的是DX9还是DX11?请楼主告诉我,我才能回答=========================================== 重装一下呢===========================================看看你显卡是否能带的动那游戏,如果能带的动 那就是显卡驱动问题 以为使命召唤8是新游戏 所有之前的驱动没有优化好也是可以理解的,去官网下最新的显卡驱动=========================================== 把显卡驱动重装。===========================================
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Powered by& 第二关按下F后出现unknow block compression type ...
查看: 1739|回复: 13
UID8696987主题阅读权限10帖子精华0积分3金钱22 荣誉0 人气0 在线时间11 小时评议0
Lv.1游侠新人, 积分 3, 距离下一级还需 2 积分
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第二关按下F后卡出来 出现unknow block compression type&&是什么原因
UID8696987主题阅读权限10帖子精华0积分3金钱22 荣誉0 人气0 在线时间11 小时评议0
Lv.1游侠新人, 积分 3, 距离下一级还需 2 积分
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自己顶一下
UID8696987主题阅读权限10帖子精华0积分3金钱22 荣誉0 人气0 在线时间11 小时评议0
Lv.1游侠新人, 积分 3, 距离下一级还需 2 积分
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不要沉下去
可爱的小花猪
夏磊游侠元老版主『动作/射击游戏区』永远萌系会员·小花猪·夏磊Assassin·真神之语·玛利亚
UID449654主题阅读权限150帖子精华0积分98758金钱55889 荣誉772 人气13953 在线时间65535 小时评议37
帖子精华0积分98758金钱55889 荣誉772 人气13953 评议37
是游戏跳出的提示？
游戏运行库工具：http://www.ali213.net/zhuanti/yxk/
UID8696987主题阅读权限10帖子精华0积分3金钱22 荣誉0 人气0 在线时间11 小时评议0
Lv.1游侠新人, 积分 3, 距离下一级还需 2 积分
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xialei_026 发表于
是游戏跳出的提示？
第二关不是那个人进棺材了 要按F致敬的时候 就卡住 跳出来了
UID8696987主题阅读权限10帖子精华0积分3金钱22 荣誉0 人气0 在线时间11 小时评议0
Lv.1游侠新人, 积分 3, 距离下一级还需 2 积分
帖子精华0积分3金钱22 荣誉0 人气0 评议0
UID1988100主题阅读权限20帖子精华0积分121金钱2107 荣誉0 人气0 在线时间391 小时评议0
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同样问题,type:216,求解?
可爱的小花猪
夏磊游侠元老版主『动作/射击游戏区』永远萌系会员·小花猪·夏磊Assassin·真神之语·玛利亚
UID449654主题阅读权限150帖子精华0积分98758金钱55889 荣誉772 人气13953 在线时间65535 小时评议37
帖子精华0积分98758金钱55889 荣誉772 人气13953 评议37
打1号升级档看看：
游戏运行库工具：http://www.ali213.net/zhuanti/yxk/
UID1988100主题阅读权限20帖子精华0积分121金钱2107 荣誉0 人气0 在线时间391 小时评议0
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我是已经打过了,可是还是一样啊!
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同样的问题，读取第二关的时候直接跳出错误对话框 unknow block compression type:250
一号补丁已经打过，待解决
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我是卡在第六关片头
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我也是类似问题，第一关开场动画以后，按完f键帮着把队又拉回舱内，然后飞行器坠落以后，直接当的一声跳出。显示unknow block compression type:40.1，已经打过了2.8g的那个（第一号补丁及游侠破解）。2，已经修改了高级系统设置里的分页文件为“系统管理的大小”。3，进游戏以前已经把输入法设定为英文输入。苦等高人出手相助！！
UID1299085主题阅读权限20帖子精华0积分20金钱333 荣誉0 人气0 在线时间71 小时评议0
Lv.2游侠会员, 积分 20, 距离下一级还需 180 积分
帖子精华0积分20金钱333 荣誉0 人气0 评议0
我也是类似问题，第一关开场动画以后，按完f键帮着把队友拉回舱内，然后飞行器坠落以后，直接当的一声跳出。显示unknow block compression type:40.
1，已经打过了2.8g的那个（第一号补丁及游侠破解）。2，已经修改了高级系统设置里的分页文件为“系统管理的大小”。
3，进游戏以前已经把输入法设定为英文输入。
苦等高人出手相助！！
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Powered byStatic analysis of an unknown compression format - LSE Blog
Categories
I really enjoy reverse engineering stuff. I also really like playing video
games. Sometimes, I get bored and start wondering how the video game I'm
playing works internally. Last year, this led me to analyze Tales of Symphonia
2, a Wii RPG. This game uses a custom virtual machine with some really
interesting features (including cooperative multithreading) in order to
describe cutscenes, maps, etc. I started to be very interested in how this
virtual machine worked, and wrote a (mostly) complete implementation of this
virtual machine in C++.
However, I recently discovered that some other games are also using this same
virtual machine for their own scripts. I was quite interested by that fact and
started analyzing scripts for these games and trying to find all the
improvements between versions of the virtual machine. Three days ago, I started
working on Tales of Vesperia (PS3) scripts, which seem to be compiled in the
same format as I analyzed before. Unfortunately, every single file in the
scripts directory seemed to be compressed using an unknown compression format,
using the magic number "TLZC".
Normally at this point I would have analyzed the uncompress function
dynamically using an emulator or an on-target debugger. However, in this case,
there is no working PS3 emulator able to help me in my task, and I also don't
possess an homebrew-enabled PS3 to try to dump the game memory. Sadface. I tend
to prefer static analysis to dynamic analysis, but I also didn't know a lot
about compression formats at this point. Still, I started working on reversing
that format statically.
I started by decrypting the main game executable (thanks, f0f!) to check if I
could find the uncompress function in the binary. Unluckily, cross-references
did not help me find anything, and immediate values search (in order to find
the FOURCC) did not lead me to anything. I was stuck with 500 compressed files
and a binary where I was not able to find the interesting code.
Oh well. Let's start by analyzing the strings in this binary:
12$ strings eboot.elf | grep -i compr
Warning: Compressed data at address 0x%08X is *bigger* than master data (%d & %d).
Pointless?
Warning: Compressed data at address 0x%08X is *bigger* than master data (%d & %d).
Pointless?
The file doesn't contain any compressed frames yet.
EDGE ZLIB ERROR: DecompressInflateQueueElement returned (%d)
unknown compression method
EDGE ZLIB ERROR: Stream decompressed to size different from expected (%d != %d)
EDGE LZMA ERROR: Size of compressed data is %d.
Maximum is %d.
EDGE LZMA ERROR: Size of uncompressed data is %d.
Maximum is %d.
EDGE LZMA ERROR: DecompressInflateQueueElement returned (%d)
*edgeLzmaInflateRawData error: compressed bytes processed (0x%x) is not value expected (0x%x)
*edgeLzmaInflateRawData error: uncompressed bytes processed (0x%x) is not value expected (0x%x)
We have references to an LZMA decompression library as well as zlib. However,
if we compare a TLZC header to some zlib'd data and an LZMA header, they do not
really look alike:
ccf0 0f00 80d4 8300
TLZC............
00 00 5d00 79
........]....R.y
e02 165c 164a 11cf 52 0052
...\.J...R.R.R.R
9c d4bd 0f7c 54d5 993f 7cee dc49 3281
x....|T..?|..I2.
dc4d 809a d8d2 76c2
X.vl..2..M....v.
32 51ec 3b48 94e9 fb0b 6bb4 b424 12bb
@2Q.;H....k..$..
d00 0080 00ff ffff ffff ffff ff00 3f91
].............?.
d923 f0 887d
E.hDTb.#t9..$..}
aabf d038 d3a0 232a
...~.&ZI...8..#*
Looking further into the TLZC header though, there is something that looks very
interesting: the 5D 00 00 01 00 string is almost like the 5D 00 00 80 00
string from the LZMA header. Looking at some
I was able to figure out that 5D is a very classic LZMA parameters value. It
is normally followed by the dictionary size (in little endian), which by
default is 0x with my LZMA encoder but seems to be 0x in
the TLZC file. The specification tells us that this value should be between
0x and 0x, so the TLZC value is in range and could be
My first try was obviously to try to reconstruct a valid LZMA header (with a
very big "uncompressed size" header field) and put it in front of the TLZC data
(header removed):
1lzma: test.lzma: Compressed data is corrupt
Sadface. Would have been too easy I guess.
Let's take a closer look at the TLZC header:
Bytes 0 - 4: FOURCC "TLZC"
Bytes 4 - 8: unknown (maybe some kind of version?)
Bytes 8 - 12: matches the size of the compressed file
Bytes 12 - 16: unknown but might be the size of the uncompressed file: for
all of the 500 files it was in an acceptable range.
Bytes 16 - 24: unknown (all zero)
Bytes 24 - 29: probably LZMA params
Stuck again with nothing really interesting. I started looking at random parts
of the file with an hexadecimal editor in order to notice patterns and compare
with LZMA. At the start of the file, just after the header, the data seem to
have some kind of regular structure that a normal LZMA file does not have:
719 131f 1f92
1b05 150b 13fd
......$.........
1bf8 1f75 c
$F...3.....u R.\
23 11a0 0fe2 149b e 1a5f 1347
.#.........^._.G
ca 213f 0e1e 8c 217d 12ee
..!?...`.`..!}..
2b 17f7 124f 1bed 21d1 095b 13e5 1457
.+...O..!..[...W
44 23ca 18f6 0c9f 1aa1 a9
.D#..........P#.
c1 160b 137c 172c 246a
.....|.,$j......
In this range there are a lot of bytes in [0x00-0x20] each followed by a byte
in [0x80-0xFF]. This is quite different from the start of a normal LZMA file,
but at that point that doesn't help us a lot.
This made me think of entropy. If I was able to measure the frequency of each
byte value in the file maybe I could compare it to some other compression
format or notice something. I created a simple Python file which counts the
occurrences of each byte value. For single byte values, this did not give any
interesting results: max count is 4200 and min count is 3900, no notable
variation, etc. However, looking at the byte digrams showed me something very
interesting: the 00 7F digram occurred 8 times more than most digrams, and
the 00 00 digram twice as much. I followed this lead and looked at what bytes
where next after 00 7F:
00 7F 9F occurs 4x more than all other digrams
00 7F 9F 0C is the only substring that starts with 00 7F 9F
Next byte is almost always C6 but in very few cases A6 also occurs
After that, the next byte is between 0x78 and 0x7C, most of the time
either 7B or 7C
No statistical evidence of a byte occurring more than the others after this.
In a 8MB file, the string 00 7F 9F 0C C6 7B occurred exactly 25 times. That's
a lot, but short strings like this do not really give us any big information. I
started to look at the 00 00 digram after that and got something a lot more
interesting: a very long repeating sequence. In the file I was analyzing, this
0x52 bytes sequence appeared 3 times in a row:
600 7fb6 f 1703 fe0f f577 302c
.............w0,
0000010: d378 4b09 691f 7d7a bc8e 340c f137 72bc
.xK.i.}z..4..7r.
..N....I.Q]..Vc.
0000030: edea b3a1 9f6d d986 34b3 f14a f52b 43be
.....m..4..J.+C.
c50 94a5 747d 40cf 85ee db27 f30d c6f7
.P..t}@....'....
I tried to discern some kind of patterns in these data, tried to grep some of
the files on my system to find parts of this sequence, no luck. Stuck again.
After a long night, I came back to my notes and remembered the start of the
file where there were strange byte patterns. If I started at offset 59 and took
2 bytes little endian integers from there, each of these integers was less than
0x2000, and often in the same range. But more interesting than that fact:
there was three consecutive integers equal to 0x0052, the size of the three
times repeating block I noticed earlier.
That's when I got the idea that ultimately solved this puzzle: TLZC files are
not one big compressed block, but several blocks, each compressed individually.
The size of these compressed blocks is contained in the header. That's actually
a very common structure used to allow semi-random access in a compressed file:
you don't need to uncompress the whole file but only the part of the file which
contains the data you want. It seemed to make a lot of sense, so I went
with it and tried to find evidence that failed my hypothesis.
If this file is indeed compressed by block, there must be somewhere in the
header either the number of blocks and their size, either the full size of the
uncompressed file and the blocks size. I went back to the TLZC header, and more
precisely to the field that I thought (without a lot of evidence) to be the
uncompressed file size. To confirm that it was it, I tried computing the
compression ratio of all of the files using the compressed size and the
uncompressed size. It gave me a plot like this:
That confirms my theory: there is a bit of noise and some files compressed a
bit more than the others, but this is still almost constant. We now have the
size of each uncompressed file, we're just missing the size of an uncompressed
If each block is independently compressed as I assumed, taking the 0x52 sized
block from above and uncompressing it should get us some results. Fail: after
adding an LZMA header, trying to uncompress the file still fails at the first
bytes of the block. Sadface again. But, thinking about it, we may know that the
block size is 0x52 but we never confirmed where exactly it started! I
generated all possible rotations of this block, and tried uncompressing each
6lzma: rot0.bin.test.lzma: Compressed data is corrupt
lzma: rot1.bin.test.lzma: Unexpected end of input
lzma: rot2.bin.test.lzma: Compressed data is corrupt
lzma: rot3.bin.test.lzma: Compressed data is corrupt
lzma: rot4.bin.test.lzma: Compressed data is corrupt
lzma: rot5.bin.test.lzma: Compressed data is corrupt
F. Yeah. We finally uncompressed something which seems valid, but now LZMA
can't find the input end marker and deletes the output file. Using strace, I
can see that the output was exactly 0x10000 bytes before it was unlinked:
7write(2, &lzma: &, 6lzma: )
write(2, &rot1.bin.test.lzma: Unexpected e&..., 43) = 43
write(2, &\n&, 1
lstat(&rot1.bin.test&, {st_mode=S_IFREG|0600, st_size=65536, ...}) = 0
unlink(&rot1.bin.test&)
Let's try putting the size in the LZMA header instead of letting the decoder
figure out the size (there is an optional "size" field in the LZMA header). As
expected, it works just fine and the uncompressed file is 0x10000 bytes long.
The data in it is obviously a bit repetitive (compressed to 52 bytes...) but
seems coherent (looks like part of an ARGB image to me):
60000000: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
0000010: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
0000020: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
0000030: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
0000040: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
0000050: ffd8 b861 ffd8 b861 ffd8 b861 ffd8 b861
...a...a...a...a
At that point I could almost uncompress the whole file, but we don't know where
the blocks data start in the file because we don't know how much blocks there
are. To test a bit more the decompressing process, I tried taking the block
just before the first 0x52 block: I can see in the block size table at the
start of the file that its size is 0x9CF, so it must start at offset 0x6415
in the file (because the 0x52 block was at offset 0x6D67). Extracting it
works too, and its size is also 0x10000. It seems to be part of the same ARGB
image (being just before, it was kind of obvious), but less repetitive this
60000000: fffe da9e fffe de9e fffa da9e ff86 7457
..............tW
0000010: ff66 3232 ffc6 5252 ffc6 5252 ffc3 5151
.f22..RR..RR..QQ
0000020: ffc3 5151 ffc3 5151 ffc3 5151 ffc3 5151
..QQ..QQ..QQ..QQ
0000030: ffc3 5151 ffc3 5151 ffc3 5151 ffc3 5151
..QQ..QQ..QQ..QQ
0000040: ffc3 5151 ffc3 5151 ffc3 5151 ffc3 5151
..QQ..QQ..QQ..QQ
0000050: ffc3 5151 ffc3 5151 ffc3 5151 ffc3 5151
..QQ..QQ..QQ..QQ
From there I uncompressed a few other blocks around the 0x52 block, and each
of these blocks was 0x10000 bytes long. I assumed that it was some kind of
constant size. From there, we can easily get the number of blocks in the file:
just take the uncompressed file size, divide it by the block size we just found
(rounding correctly!) and here is your number of blocks!
For the first file, uncompressed size is 8639616, which means 132 blocks
are required. This means that the first block data is at offset:
1header_size (1D) + number_of_blocks * sizeof (uint16_t)
Uncompressing that first block gives us something interesting that validates
everything we've done so far:
650 06 00 0080
FPS4............
10 00 00 0080
...G............
00 f280 00
................
................
7c 6c00 007c 6c00 7d 6d00
.|l..|l......}m.
00 9a 7d 7300
.............}s.
The FPS4 FOURCC is an archiving format commonly used in Tales of games.
That means we actually uncompressed valid stuff and not only garbage!
From there, it's easy to write a full decompression software. Here is mine,
written in Python using PyLZMA:
40import mmap
import pylzma
import struct
import sys
UNCOMP_BLOCK_SIZE = 0x10000
def decompress_block(params, block, out, size):
block = params + block
out.write(pylzma.decompress(block, size, maxlength=size))
def decompress_tlzc(buf, out):
assert(buf[0:4] == &TLZC&)
comp_size, uncomp_size = struct.unpack(&&II&, buf[8:16])
num_blocks = (uncomp_size + 0xFFFF) / UNCOMP_BLOCK_SIZE
lzma_params = buf[24:29]
block_header_off = 29
data_off = block_header_off + 2 * num_blocks
remaining = uncomp_size
for i in xrange(num_blocks):
off = block_header_off + 2 * i
comp_block_size = struct.unpack(&&H&, buf[off:off+2])[0]
block = buf[data_off:data_off+comp_block_size]
data_off += comp_block_size
if remaining & UNCOMP_BLOCK_SIZE:
decompress_block(lzma_params, block, out, remaining)
decompress_block(lzma_params, block, out, UNCOMP_BLOCK_SIZE)
remaining -= UNCOMP_BLOCK_SIZE
if __name__ == &__main__&:
fd = os.open(sys.argv[1], os.O_RDONLY)
buf = mmap.mmap(fd, 0, prot=mmap.PROT_READ)
decompress_tlzc(buf, open(sys.argv[2], &w&))
os.close(fd)
Three days of work for 40 lines of Python. So worth it!
This was a very interesting experience for someone like me who did not know a
lot about compression formats: I had to look a lot into LZMA internals, read
tons of stuff about how it works to try to find some patterns in the compressed
file, and found out that the most simple methods (trying to find repeating
substrings) give actually a lot of interesting results to work with. Still, I
hope next time I work on such compression algorithm I'll have some code to work
with or, even better, an emulator!
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