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2009年1月13日
during analysis of "IO Error: Connection reset", many articles mentioned that it could be caused by java security code (accessing /dev/random) used in JDBC connection. However it is not the root cause in my case.
In my environment, Java already use /dev/urandom.
1. $JAVA_HOME/jre/lib/security/java.security
securerandom.source=file:/dev/./urandom
2. check with strace.
only -Djava.security.egd=file:/dev/../dev/urandom will trigger system call (read on /dev/urandom)
all other other path format like below are OK.
-Djava.security.egd=file:/dev/./urandom
-Djava.security.egd=file:///dev/urandom
3. Keep checking the retropy size, I have never seen it is exhaused.
while [ 1 ];
do
cat /proc/sys/kernel/random/entropy_avail
sleep 1
done
usually the avail is in the range from 1000 to 3000.
so far, there is no clue about the root cause of "IO Error: Connection reset".
I encountered many issue during installation of Oracle Grid Infrastructure(GI) and Database;
with the help of ariticle and documents found through Google search engine,
I finally made it. for records, here is the details issues encountered and solutions applied.
Major issues were encountered during GI installation.
Pre-installation tasks.
Issue 1: swapspace is not big enough; (1.3.1 Verify System Requirements)
grep MemTotal /proc/meminfo
264G
grep SwapTotal /proc/meminfo
2G
during OS installation, I take default option and swap space is only 2G.
Oracle recommend to have more than 16G swap space in case of more that 32G RAM.
dd if=/dev/zero of=/home/swapfile bs=1024 count=33554432
33554432+0 records in
33554432+0 records out
34359738368 bytes (34 GB) copied
mkswap /home/swapfile
mkswap /home/swapfile
chmod 0600 /home/swapfile
lessons learned: setup swap space properly according to DB requirement when installing OS.
Issue 2: cannot find oracleasm-kmp-default from Oracle site.
(1.3.6 Prepare Storage for Oracle Automatic Storage Management)
install oracleasmlib and oracleasm-support is easy, just download them from Oracle and install them;
Originally oracleasm kernel is provided by Oracle, but now I cannot find it from Oracle; finally I
realized that oracleasm kernel is now provided by OS vendor;
In my case, it should be installed from SUSE disk;
a. to get its name oracleasm-kmp-default
zypper se oracle
b. map dvd and install
zypper in oracleasm-kmp-default
rpm -qa|grep oracleasm
oracleasm-kmp-default-2.0.8_k3.12.49_11-3.20.x86_64
oracleasm-support-2.1.8-1.SLE12.x86_64
oracleasmlib-2.0.12-1.SLE12.x86_64
asm configure -i
asm createdisk DATA /dev/<...>
asm listdisks
--DATA
ls /dev/oracleasm/disks
Installation tasks:
Issue 3: always failed due to user equivalence check after starting installer OUI with user oracle.
however if I manully check with runcluvfy, no issue found at all.
./runcluvfy.sh stage -pre crsinst -n , -verbose
I worked around it by using another user to replace user oracle. but it triggered next issue.
Issue 4: cannot see ASM disks in OUI. no matter how I change the disk dicovery path. the disk list is empty.
but I can find disk manully.
/usr/sbin/oracleasm-discover 'ORCL:*'
Discovered disk: ORCL:DATA
Root cause is that the ASM is configured and created with user oracle. and I aming installing GI
with different user other than oracle; so I cannot see the Disk created.
change owner of disk device file solved the issue.
ls /dev/oracleasm/disks
chown /dev/oracleasm/disks -R
Issue 5: root.sh execution failed.
Failed to create keys in the OLR, rc = 127, Message:
clscfg.bin: error while loading shared libraries: libcap.so.1:
cannot open shared object file: No such file or directory
fixed the issue with command below:
zypper in libcap1
ohasd failed to start
Failed to start the Clusterware. Last 20 lines of the alert log follow:
2016-07-24 23:10:28.502:
[client(1119)]CRS-2101:The OLR was formatted using version 3.
I found a good document from SUSE,
Oracle RAC 11.2.0.4.0 on SUSE Linux Enterprise Server 12 - x86_64,
it make it clear that SUSE 12 is supported by Oracle GI 11.2.0.4, it also mentioned
Patch 18370031.
"During the Oracle Grid Infrastructure installation,
you must apply patch 18370031 before configuring the software that is installed. "
The patch 18370031 is actually mentioned in "Oracle quick installation guide on Linux",
but not mentioned in "Oracle quick installation guide on Linux". I majored followed up
with later one and missed Patch 18370031.
issue disappeared after I installed the patch 18370031.
./OPatch/opatch napply -oh -local /18370031
Errors in file :
ORA-27091: unable to queue I/O
ORA-15081: failed to submit an I/O operation to a disk
ORA-06512: at line 4
solved by change owner of disk DATA related file
ls -l /dev/oracleasm/iid
chown on folder /dev/oracleasm/iid and some .* hidden file.
Issue during DB installation
Issue 6: report error: in invoking target 'agent nmhs'
vi $ORACLE_HOME/sysman/lib/ins_emagent.mk
Search for the line
$(MK_EMAGENT_NMECTL)
Change it to:
$(MK_EMAGENT_NMECTL) -lnnz11
refer to
https://community.oracle.com/thread/1093616?tstart=0
很多年前裝了Ubuntu和Windows的雙系統，最近因為有了專門的電腦來裝Ubuntu，所以把原先電腦上的Ubuntu卸載了，結果系統不能引導了。因為GRUB的原理是控制權從MBR到Ubuntu系統盤，然后Ubuntu系統盤再提供對windows的引導。現在Ubuntu系統被卸載了。這個啟動的鏈條也就斷了。
這個問題本身不難解決，借了一個Windows安裝盤，恢復一下MBR即可。但是這個需要windows系統的Administrator密碼。而我的系統不是我裝的，我根本不知道這個密碼。
有的帖子提到破解Administrator密碼，試了一下，覺得太麻煩了。因為電腦上有數據，也不能重裝。
最后的解決方案是在原先的Ubuntu分區上安裝一個新的Windows。這樣變成了windows的雙系統。安裝完重啟之后可以進入任何一個系統（新的或者舊的Windows）。安裝的過程中MBR被自動更新了。再下來就改一下老系統的Administrator密碼，刪除掉多于的Windows新系統即可。
用Gmail的時候不小心點了"存檔"按鈕，一封重要的郵件就此消失了好幾天，今天才機緣巧合找到。
在網絡上查到的解釋是：
存檔會將郵件從收件箱移至所有郵件，這樣您不必刪除郵件就可以整理收件箱。
難以理解，坦率地說，這個功能對我來說是徒增煩惱。看來任何工具都需要你去適應，磨合。
最近買了一個叫做“華容道”的玩具給兒子晚。這個游戲雖然號稱是中國四大古典智力游戲之一。其實不過百年歷史，而且是從國外引進的。不過本地化做得非常好，也算是創造性地吸收國外文明。
手工解決這個游戲有點難度，當然已經有人給出了解法；不過我還是自己用編程的方式解決了一遍。發現自己在這方面的編程還是比較弱。大部分時間花在了調試上。
剛開始是用的深度優先搜索。大致知道了答案應該長什么樣。后來改進為廣度優先搜索，得到了最優的解法。還有一個就是原先只考慮每次最多移動一格。后來發現傳統的定義是一個塊的所有連續移動都算作一步。相應地修改了實現算法。
最難的是做界面。為了調試，隨便寫了個Applet。但是給我兒子玩，就覺得拿不出手了。
Just use this blog to share some meta information. git://github.com/ueddieu/mmix.git http://github.com/ueddieu/mmix.git
After two weeks' struggle, I have successfully installed Gentoo, a popular GNU/Linux Distribution. For Records, the obstacles I encountered are listed below.
(but I can not remember the solution exactly)
0. failed to emerge gpm when I install the links package.
If I recall correctly, it is resolved by install gpm manually 1. I encounter issue when I install glib 2.22.5.
no update-desktop-database.
which is in dev-util/desktop-file-utils. When I try to emerge it, there is a circular dependency on glib. no solution
and I forget How I resolve the problem.
2. later after I install glib, with ~amd64 keyword I can install gpm-1.20.6, but it conflicts with the manually inatalled gpm.
I remove the conflicted file and emerge successfully.
3. Failed to emerge tiff.
edit packages.keywords to add the following.
/ ~amd64
I am able to use latest tiff in beta-version, which is unstable and masked out.
4. later atk-1.28.0 failed to emerge.
edit /etc/make.conf with the following.
FEATURES="-stricter".
then emerge successfully with only some complain. with out this seting. the warining from GCC will cause that emerge fail.
5. when I run
emerge --update system
actually gcc will be upgraded from 4.3.4 to 4.4.3. but it failed because of compilation warning, again. add "-stricter" into Features variable in /etc/make.conf work around it.
6. The installation takes a long time, the KDE itself take more than 10 hours. There is still a lot of improvement space! Anyway, it is nice to be able to use it daily.
在C:\Documents and Settings\<user_name>\Application Data\Subversion\servers文件中加入
all=*.*
[all]
http-proxy-host = ***.**.com
http-proxy-port = 8080
這里的all映射到所有的Server。
網絡環境的復雜給我們的工作帶來了一些影響。就拿Proxy的設置來說，本來理想的情況是在全局做一個設置就可以了，但是事實上我們要為每個程序做設置，而且語法還不一樣。
今天從word文檔中拷貝腳本到命令行執行。沒有想到的是Word自動加入了空格，導致執行失敗。具體如下。
call ttGridCreate('$TT_GRID');
被word變成了
call ttGridCreate(' $TT_GRID');
這個空格可不容易被發現，尤其你不是腳本作者的時候。提高警惕！
今天做了一個簡單的性能測試。比較訪問Java對象屬性的各種方法的性能差異。
1. 直接訪問對象的屬性。
2. 用方法訪問對象的屬性。
3. 用Map來存儲和訪問。
4. 反射－Field 訪問。
5. 反射－Method訪問。
重復100次，結果如下(單位為納秒）。
* 100 field access, 14,806<br/>
* 100 method access, 20,393<br/>
* 100 map access, 66,489<br/>
* 100 reflection field access, 620,190<br/>
* 100 reflection method access, 1,832,356<br/>
重復100000次，結果如下(單位為納秒）。
*100000 field access, 2,938,362
*100000 method access, 3,039,772
*100000 map access, 10,784,052
*100000 reflection field access, 144,489,034
*100000 reflection method access, 37,525,719 <br/>
由結果可見：
1。getter/setter 的性能已經接近直接屬性訪問（大約慢50％），沒有必要擔心getter/setter的性能而采用直接屬性訪問。
2。用Map代替POJO的代價大約是比getter/setter慢三倍。
3。反射訪問比getter/setter慢50到150倍。慎用。追求動態性的時候也要注意不菲的性能代價。
4。注意重復次數增加到100000次，方法訪問和屬性訪問的差距縮小；更有意思的是，反射的Method訪問比Field訪問快四倍。這主要是JIT的作用。
該測試結果和原先的猜想基本符合。但是性能評估很容易得到片面的結論，如果有錯誤的地方，請大家不吝指正。謝謝。
0. I am reading the source code of Tomcat 6.0.26. To pay off the effort,
I documents some notes for record. Thanks for the articles about Tomcat
source code, especially the book <<How Tomcat works>>.
1. They are two concepts about server, one is called Server, which
is for managing the Tomcat (start and stop); another is called Connector,
which is the server to serve the application request. they are on the different
ports. The server.xml clearly show the difference.
<Server port="8005" shutdown="SHUTDOWN">
<Service name="Catalina">
<Connector port="8080" protocol="HTTP/1.1"
connectionTimeout="20000"
redirectPort="8443" />
<Connector port="8009" protocol="AJP/1.3" redirectPort="8443" />
although the server is the top level element, logically it should not be.
Actually in code, Bootstrap starts the service first, which
in turn start the Server and server's services.
2. My focus in on Connector part. I care how the request is services by the
Tomcat. Here are some key classes.
Connector --> ProtocolHandler (HttpProtocol
and AjpProtocol) --> JIoEndPoint
--> Handler(Http11ConnectionHandler
and AjpConnectionHandler)
3. Connector is most obervious class, but the entry point is not here.
The sequence is like this.
Connector.Acceptor.run()
--> JioEndPoint.processSocke(Socket socket)
-->SockeProcess.run()
-->Http11ConnectorHandler.process(Socket socket)
-->Http11Processor.process(Socket socket)
-->CoyoteAdapter.service(Request req, Response res)
The core logic is in method Http11Processor.process(Socket socket)
CoyoteAdapter.service(Request req, Response res) bridges between Connector module and Container module.
Any comments are welcome. I may continue the source code reading and dig deeper into it if time permit.
It is handy to be able to navigate the source code with Ctrl + ] in Cscope, but I always forget how to navigate back and waste effort many times. So for record, Ctrl+t can navigate back in Cscope.
One more time, Ctrl+] and Ctrl+t can navigate forth and back in Cscope.
How to read the source code in <<TCP/IP Illustrated Volume 2>>
1. Get the source code, original link provided in the book is not available now.
You may need to google it.
2. install cscope and vi.
3. refer to http://cscope.sourceforge.net/large_projects.html for the following steps.
It will include all the source code of the whole OS, not only the kernel.
find src -name '*.[ch]' > cscope.files
we actually only care kernel source.
find src/sys -name '*.[ch]' > cscope.files
4. wc cscope.files
1613 1613 45585 cscope.files
5. vim
:help cscope
then you can read the help details.
6. if you run vim in the folder where cscope.out resides. then it will be loaded
automaically.
7. Try a few commands.
:cs find g mbuf
:cs find f vm.h
They works. A good start.
P.S. this book is quite old, if you know it well and can recommend some better alternative for learning TCP/IP, please post a comments, Thanks in advance.
我兒子在彈鋼琴，他阿姨說，“哥哥就喜歡彈難的。”
我外甥女說：“哥哥要彈男的。妹妹要彈女的。”
在高中的時候，知道了用篩法可以得到素數。當時我還有一個錯誤的關于尋找素數的猜測。
以為用兩個素數相乘，其附近存在素數的幾率很高。比如， 7×11 ＝ 77， 其附近有79，正好是素數。
當時已經發現11×11＝121。7×17＝119；但是錯誤的理解為只有其中一個是平方或次冪時才成立。
后來有了計算機，編程驗證了一下，發現有很多的反例。對當初的錯誤猜測羞赧不已。
這個猜測雖然錯的離譜，但是和現在的素數理論，尤其是孿生素數還是很有關系的。現在已經知道，
素數有無窮多個，但是素數在自然數中所占的比例逐漸趨近于零。
因此孿生素數在自然數中的比例也是趨近于零的。現在還沒有證明孿生素數是否有無窮多個。
這個猜測的樸素之處在于，任何兩個素數之乘積A，要么A是3n+2，要么A是3n+1；如果是3n+2，則只有A+2
才有可能是素數；如果是3n+1，則只有A-2才有可能是素數。但是，事實上，這個猜測成立的比例非常的低。
寫了一個程序驗證了一下。16位的整數中，大概只有 10％ 能使假設成立。
由于是在Proxy的網絡環境，MSYSGIT 的 git clone 總是失敗。需要配置如下環境變量。
export http_proxy="http://<proxy domain name>:<port>"
之后http協議git clone沒有任何問題。但是用git 協議仍舊有問題。
之后發現git push 和 git pull 經常不能work。多次嘗試后發現用更全的命令行參數可以解決問題。
過程如下。
git pull --fail
git pull origin --fail
git pull git@github.com:ueddieu/mmix.git --it works.
It seems the Command line short cuts are lack of some user information, such as user name "git".
(which is kind of strange at the first glance.)
git push --fail
git push origin --fail
git push git@github.com:ueddieu/mmix.git master --it works.
Anyway, now I can check in code smoothly. :)
There are a few cases in which the un-visible blank character will cause
problem, but it is hard to detect since they are not visible.
One famous case is the '\t' character used by Make file, it is used to mark
the start of a command. If it is replace by blank space character, it does
not work, but you can not see the difference if you only look at the make file.
This kind of problem may get the newbies crazy.
Last week, I have encounter a similar issue, which is also caused by unnecessary
blank space.
As you may know, '\' is used as line-continuation when you have a very long line, e.g.
when you configure the class path for Java in a property file, you may have something like this.
classpath=/lib/A.jar;/lib/B.jar;\
/lib/C.jar;/lib/D.jar;\
/lib/E.jar;/lib/log4j.jar;\
/lib/F.jar;/lib/httpclient.jar;
But if you add extra blank space after the '\', then you can not get the complete
content of classpath. Because only when '\' is followed by a '\n' on Unix or '\r''\n'
on Windows, it will work as line-continuation ; otherwise, e.g '\' is followed by
' ''\n', the line is complete after the '\n', the content after that will be the start of
a new line.
Fortunately, it is easy to check this kind of extra blank space by using vi in Unix.
use command '$' to go to the end of line, if there is no extra blank space after '\',
the current position should be '\', if there are any blank space after '\', the current position
is after the '\'.
媽媽和兒子
媽媽，你最近不吃魚，變笨了吧――2009.6.2
兒子又要求錄音了。我們按著臺詞在dialog。
“……”兒子
“……，I like mangoes”媽媽
“媽媽，我昨天剛教會你，又忘了？是I like watermelon。”
“哦，媽媽現在記性不好了！”
“是你這幾天不吃魚了吧，變笨了吧。明天多吃點！”
媽媽你穿這衣服蠻可愛――2009-6-9晚
兒子挑選的故事講完了。“好，ok，我們睡覺吧！”我說。
“唉，媽媽，還沒錄音呢，我去拿mp3。”自從第一次提議給他錄音，兒子每天都要求我能做到。
……
“……，媽媽，你蠻可愛的！……”錄音正起勁，兒子突然插了一句題外話。
“什么？”我沒聽清。
“你穿這衣服蠻可愛的！”兒子賊賊的笑著又重復了一遍。“因為你的衣服象小斑馬呀！”我終于明白。
我安裝openldap時主要是參考了http://hexstar.javaeye.com/blog/271912
我遇到的一個新問題是執行ldapsearch報錯如下：
can not find libdb-4.7.so.
我的解決辦法是，建立符號鏈接/usr/lib/libdb-4.7.so， 后者指向/usr/local/BerkeleyDB/lib/libdb-4.7.so
之后沒有遇到其他問題。
凡事都有其內在原因，即使是表面上毫無道理的行為，也有其內在的原因。今天再次認識到這個
道理，還是因為今天早上和我兒子的一段插曲。
今天早上，我兒子不肯起床，在床上哭鬧，不讓他媽媽去上班，要他媽媽陪他睡覺。
他媽媽要趕班車，沒時間陪她，留我在家里。我陪他睡了一會，聊了十分鐘，才知道他是有原因
的。
昨天晚上，我和她媽媽都很累，我就說今天我們早點睡覺，和兒子一起睡好了。可是因為剛剛
回上海，有很多事情要做，最后還是忙到十點半才睡。兒子就說爸爸說謊了。當然他可能還有
其它的原因，比如想我們每天和他一起睡覺。
The analysis of MOR(MXOR) instruction implementation in MMIXWare
-- A stupid way to understand the source code.
the implementation of MOR(MXOR) is in file: mmix-arith.w
436 octa bool_mult(y,z,xor)
437 octa y,z; /* the operands */
438 bool xor; /* do we do xor instead of or? */
439 {
440 octa o,x;
441 register tetra a,b,c;
442 register int k;
443 for (k=0,o=y,x=zero_octa;o.h||o.l;k++,o=shift_right(o,8,1))
444 if (o.l&0xff) {
445 a=((z.h>>k)&0x01010101)*0xff;
446 b=((z.l>>k)&0x01010101)*0xff;
447 c=(o.l&0xff)*0x01010101;
448 if (xor) x.h^=a&c, x.l^=b&c;
449 else x.h|=a&c, x.l|=b&c;
450 }
451 return x;
452 }
It takes me several hours to understand the details.
If we treat each octabyte as a matrix, each row corresponds to a byte, then
y MOR z = z (matrix_mulitiply) y
For a=((z.h>>k)&0x01010101)*0xff;
(z.h>>k)&0x01010101 will get the four last bit in (z.h>>k). depends on the bit in last row,
((z.h>>k)&0x01010101)*0xff will expand the bit (either 0 or 1) into the whole row.
e.g.
ff
* 0x01010101
---------------
= ff
ff
ff
ff
----------------
= ffffffff
(depending on the last bit in each row of z, the result could be #ff00ff00. #ff0000ff, etc.)
similarily, b=((z.l>>k)&0x01010101)*0xff; will expand the last bit in each byte into the
whole byte.
over all, after these two step, the z becomes the replication of it's last row, since k vary
from 0 to 7, it will loop on all the rows actually.
For c=(o.l&0xff)*0x01010101, it will get the last byte in o.l and populate it to other three byte.
since it will not only or/xor h but also l. it is not necessary populate it to o.h.
one example,
let (z.h>>k)&0x01010101 = 0x01000101, then a= 0xff00ffff;
let (z.l>>k)&0x01010101 = 0x01010001, then b= 0xffff00ff;
let (o.l&0xff)=0xuv, then c= 0xuvuvuvuv;
then a&c=0xuv00uvuv;
b&c=0xuvuv00uv;
consider the elements [i,j] in result x. in this round, what value was accumalated in by operation
or(xor).
it is the jth bit in last byte of o.l & ith bit in last column of z.(do not consider looping now.)
in this round, the 64 combination of i and j, contirbute the value to the 64 bits in z.
Noticed that o loop on y from last byte to first byte. There are 8 loop/rounds, in another round.
say kth round.
the elements[i,j] will accumuate the jth bit in last (k + 1)th row & the jth bit in last (k+1)th
column.
that means the jth column in y multiply the ith row in z. it conform to the definiton for
z matrix_multiply y.
游戲和數學有密切的聯系。最近在玩九連環，感受更深。
之所以開始玩九連環，是因為在高德納的書中提到了格雷碼和九連環的關系。為了理解生成格雷碼的算法，特意買了九連環來玩。畢竟書上的
描述沒有實際玩起來那么容易理解。
通過這個游戲，我不僅會解九連環了，而且掌握的生成格雷碼的一種算法。
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