Django絕對(duì)簡(jiǎn)明手冊(cè)

http://wiki.woodpecker.org.cn/moin/DjangoZipManual

Django 是一個(gè)高級(jí) Python web framework，它鼓勵(lì)快速開發(fā)和干凈的、MVC設(shè)計(jì)。它包括一個(gè)模板系統(tǒng)，對(duì)象相關(guān)的映射和用于動(dòng)態(tài)創(chuàng)建管理界面的框架。Django遵守BSD版權(quán)。

Javascript 絕對(duì)簡(jiǎn)明手冊(cè)
http://wiki.woodpecker.org.cn/moin/jsInAWord

Javascript和C++,Java,Python一樣是一種博大精深的編程語言.

Docbook美化Css完美版,Css就在Css文件夾中

點(diǎn)擊下載

Docbook用來生成文檔不錯(cuò),不過寫起來太煩.

幸好http://wiki.woodpecker.org.cn/的wiki可以把wiki文檔轉(zhuǎn)換為Docbook的代碼,而寫Wiki格式的文檔就舒服多了.

我做了這個(gè)Css是為了讓Docbook生成的文檔讀起來舒服一點(diǎn)
大家可以配合CDBE(Chinese DocBook Environment（CDBE）)使用,寫文檔也很享受:)
http://manual.vingel.com/docs/data/20051013100319/index.html
Boost.Asio是利用當(dāng)代C++的先進(jìn)方法,跨平臺(tái),異步I/O模型的C++網(wǎng)絡(luò)庫
現(xiàn)在完成了的小節(jié)

?? 1. 網(wǎng)絡(luò)庫:VC2005注意
?? 2. 同步Timer
?? 3. 異步Timer
?? 4. 回調(diào)函數(shù)的參數(shù)
?? 5. 成員函數(shù)作為回調(diào)函數(shù)
?? 6. 多線程回調(diào)同步


文章見
http://wiki.woodpecker.org.cn/moin/Boost

文章是用wiki寫的，有不妥大家可以直接改正，謝謝。 找算法,有數(shù)據(jù)庫

GOOGLE上的算法太多太雜,
在專門的數(shù)據(jù)庫中找就方便多了.
我在學(xué)校是可以全文下載的.

http://portal.acm.org/portal.cfm
ACM（Association for Computing Machinery，美國(guó)計(jì)算機(jī)學(xué)會(huì)）創(chuàng)立于1947年，目前提供的服務(wù)遍及100余個(gè)國(guó)家，會(huì)員人數(shù)達(dá)80,000多位專業(yè)人士，并于1999年起開始提供電子數(shù)據(jù)庫服務(wù)――ACM Digital Library全文數(shù)據(jù)庫。ACM Digital Library全文數(shù)據(jù)庫，收錄了美國(guó)計(jì)算機(jī)協(xié)會(huì)（Association for Computing Machinery）的各種電子期刊、會(huì)議錄、快報(bào)等文獻(xiàn)，由iGroup Asia Pacific Ltd代理。2002年10月21日iGroup公司在清華大學(xué)圖書館建立了鏡像服務(wù)器。目前，我校可以訪問國(guó)內(nèi)站點(diǎn)和美國(guó)主站點(diǎn)。


http://ieeexplore.ieee.org
IEEE/IEE Electronic Library（IEL）數(shù)據(jù)庫提供美國(guó)電氣電子工程師學(xué)會(huì)（IEEE）和英國(guó)電氣工程師學(xué)會(huì)（IEE）出版的229種期刊、8739種會(huì)議錄、1646種標(biāo)準(zhǔn)的全文信息。

????? IEEE和IEE是世界知名學(xué)術(shù)機(jī)構(gòu)。IEL數(shù)據(jù)庫包含了二者的所有出版物，其內(nèi)容計(jì)算機(jī)、自動(dòng)化及控制系統(tǒng)、工程、機(jī)器人技術(shù)、電信、運(yùn)輸科技、聲學(xué)、納米、新材料、應(yīng)用物理、生物醫(yī)學(xué)工程、能源、教育、核科技、遙感等許多專業(yè)領(lǐng)域位居世界第一或前列。

????? IEL更新速度很快，一般每周更新一次，每月增加25,000篇最新文獻(xiàn)。而且，每年IEEE還會(huì)有新的出版物加入到IEL中去。 一道據(jù)說是Google的面試題
題目:有一個(gè)整數(shù)n,寫一個(gè)函數(shù)f(n),返回0到n之間出現(xiàn)的"1"的個(gè)數(shù)。比如f(13)=6 ; f(11)=4,算出f(n)=n的n(如f(1)=1)？
我用python寫了一份代碼,還改寫了一份c++代碼

python源代碼
def count(i):
"""count form 0 to this number contain how many 1
1.you shoul know pow(10,n)-1 contain 1 number is n*pow(10,n-1)
2.use 32123 for example:
from 10000 to 32123 the first digit contain 1 number is 1(0000-9999) = pow(10,4) ,
and from 10000 to 30000 the rest digit contain 1 number is ( firstDigit*4*pow(10,4-1) )
so count(32123)=pow(10,4)+( firstDigit*4*pow(10,4-1) ) + count(2123)

print count(1599985)
1599985

print count(8)
1
"""
if i==0:
return 0
if 9>=i>=1:
return 1
digit=len(str(i))
firstDigit=int(str(i)[0])
if firstDigit>1:
now=pow(10,digit-1)
else:
now=int(str(i)[1:])+1
now+=(digit-1)*pow(10,digit-2) * firstDigit
return now+count(int(str(i)[1:]))

def little(i):
count_i=count(i)

if i<count_i:

#i reduce 1 , count_i at most reduce 9 ,
#so you at least reduce (count_i-i)/(9-1) to reach i==count_i
#用位數(shù)更快
if (count_i-i)/8>1:
return i-(count_i-i)/8

if i>count_i:
#i reduce 1 , count_i at least reduce 0 , so you at least reduce (i-count_i) to reach i==count_i
return count_i

return i-1

def run(i=10*10**(10-1)):
while i>0:
# print i,'=>i-count_i= ',i-count(i)
if i==count(i):
print i,','

i=little(i)

def fastrun(t=10*10**(10-1)):
"""This just list out all this king of element :) But it is fastest and most useful"""
all=[1, 199981, 199982, 199983, 199984, 199985, 199986, 199987, 199988, 199989, 199990, 200000, 200001, 1599981, 1599982, 1599983, 1599984, 1599985, 1599986, 1599987, 1599988, 1599989, 1599990, 2600000, 2600001, 13199998, 35000000, 35000001, 35199981, 35199982, 35199983, 35199984, 35199985, 35199986, 35199987, 35199988, 35199989, 35199990, 35200000, 35200001, 117463825, 500000000, 500000001, 500199981, 500199982, 500199983, 500199984, 500199985, 500199986, 500199987, 500199988, 500199989, 500199990, 500200000, 500200001, 501599981, 501599982, 501599983, 501599984, 501599985, 501599986, 501599987, 501599988, 501599989, 501599990, 502600000, 502600001, 513199998, 535000000, 535000001, 535199981, 535199982, 535199983, 535199984, 535199985, 535199986, 535199987, 535199988, 535199989, 535199990, 535200000, 535200001, 1111111110]
for i in all:
if(t>=i):
print i

print "first test with run() to:111111111"
run(111111111)

print '>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>'
print "2st test with run() to:10^10"
run()

print '>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>'
print "3st test with fastrun() to:10^10 , Fastest!!!"
fastrun()


C++代碼
-------------------------------------------------------------------------------
#include <cmath>
#include <iostream>

using namespace std;
unsigned long long mypow(int a,int b)
{
unsigned long long sum=1;
for(int i=0;i<b;i++)
sum*=a;
return sum;
}
unsigned long long count(unsigned long long i){
/*
count form 0 to this number contain how many 1
1.you shoul know pow(10,n)-1 contain 1 number is n*pow(10,n-1)
2.use 32123 for example:
from 10000 to 32123 the first digit contain 1 number is 1(0000-9999) = pow(10,4) ,
and from 10000 to 30000 the rest digit contain 1 number is ( firstDigit*4*pow(10,4-1) )
so count(32123)=pow(10,4)+( firstDigit*4*pow(10,4-1) ) + count(2123)
*/
if(i==0)return 0;
if(9>=i and i>=1)return 1;
int digit=1;
unsigned long long firstDigit=i;
while(firstDigit>=10){
firstDigit/=10;
++digit;
}

unsigned long long now;
unsigned long long rest=static_cast<unsigned long long int>(i-(firstDigit*mypow(10,digit-1)));
if(firstDigit>1)now=static_cast<unsigned long long int>(mypow(10,digit-1));
else{now=rest+1;}
now+=static_cast<unsigned long long int>((digit-1)*mypow(10,digit-2) * firstDigit);


return (now+count(rest));
}

unsigned long long little(unsigned long long i)
{
unsigned long long count_i=count(i);

if(i<count_i){

//i reduce 1 , count_i at most reduce 9 , so you at least reduce
//用位數(shù)更快
(count_i-i)/(9-1) to reach i==count_i
if ((count_i-i)/8>1)return i-(count_i-i)/8;
}

if(i>count_i){
//i reduce 1 , count_i at least reduce 0 , so you at least reduce (i-count_i) to reach i==count_i
return count_i;
}
return i-1;
}

void run(unsigned long long i=pow(10.0f,10)){
while (i>0){
// print i,'=>i-count_i= ',i-count(i)
if(i==count(i))cout<<i<<"=>"<<count(i)<<'\n';

i=little(i);
}
cout<<"run() finished\n\n";
}

void fastrun(unsigned long long t=pow(10.0f,10)){
//This just list out all this king of element :) But it is fastest and most useful
const unsigned long long all[]={1, 199981, 199982, 199983, 199984, 199985, 199986, 199987, 199988, 199989, 199990, 200000, 200001, 1599981, 1599982, 1599983, 1599984, 1599985, 1599986, 1599987, 1599988, 1599989, 1599990, 2600000, 2600001, 13199998, 35000000, 35000001, 35199981, 35199982, 35199983, 35199984, 35199985, 35199986, 35199987, 35199988, 35199989, 35199990, 35200000, 35200001, 117463825, 500000000, 500000001, 500199981, 500199982, 500199983, 500199984, 500199985, 500199986, 500199987, 500199988, 500199989, 500199990, 500200000, 500200001, 501599981, 501599982, 501599983, 501599984, 501599985, 501599986, 501599987, 501599988, 501599989, 501599990, 502600000, 502600001, 513199998, 535000000, 535000001, 535199981, 535199982, 535199983, 535199984, 535199985, 535199986, 535199987, 535199988, 535199989, 535199990, 535200000, 535200001, 1111111110};
for(int i=0;i!=83;++i){
if(t>=all[i])cout<<all[i]<<'\n';
}
cout<<"fastrun() finished\n\n";
}
int main(int argc, char *argv[])
{
for(;;)
{
unsigned long long i;
cout<<"please input a number:";
cin>>i;
cout<<"run():\n";
run(i);
cout<<"fastrun():\n";
fastrun(i);
}
system("PAUSE");
return EXIT_SUCCESS;

}      摘要:  http://wiki.woodpecker.org.cn/moin/PyAbsolutelyZipManual最新版 PyAbsolutelyZipManual Python 絕對(duì)簡(jiǎn)明手冊(cè) -- zsp007@gmail.com ::-- ZoomQuiet [2006-09-15 04:35:33] Py2.5 絕對(duì)簡(jiǎn)明手冊(cè) ...  閱讀全文  Bjam簡(jiǎn)明教程

Bjam是Boost中一部分,但可以單獨(dú)下載,我個(gè)人覺得比make方便.

單獨(dú)下載地址
http://sourceforge.net/project/showfiles.php?group_id=7586&package_id=80982

單獨(dú)下載Bjam后,設(shè)置環(huán)境變量BOOST_BUILD_PATH到解壓目錄.

然后要在中user-config.jam選擇編譯器(就是把注釋去掉),
比如
#  Configure gcc (default version)
#     using gcc ;
改為
#  Configure gcc (default version)
     using gcc ;

在Jamroot文件中可以定義要編譯的文件和輸出的文件名的target.
如:
exe hello : hello.cpp ;
exe hello2 : hello.cpp ;

可以只編譯特定的target,如
bjam hello2

bjam可以選擇編譯版本,如
bjam debug release
bjam release

可以清理指定的版本/target
bjam --clean debug release
bjam --clean hello2

可以指定一些編譯方式
bjam release inlining=off debug-symbols=on

可以指定保含頭文件的目錄
exe hello
    : hello.cpp
    : <include>boost <threading>multi
    ;

可以為整個(gè)工程指定頭文件
project
    : requirements <include>/home/ghost/Work/boost <threading>multi
    ;

exe hello : hello.cpp ;
exe hello2 : hello.cpp ;

Boost.Asio 0.37教程 Timer.1(翻譯自Boost.Asio 0.37的文檔)

原文http://asio.sourceforge.net/boost_asio_0_3_7/libs/asio/doc/

翻譯:張沈鵬 http://blog.csdn.net/zuroc or http://m.shnenglu.com/zuroc

Timer.1 - 同步Timer
本章介紹asio如何在定時(shí)器上進(jìn)行阻塞等待(blocking wait).

實(shí)現(xiàn),我們包含必要的頭文件.

所有的asio類可以簡(jiǎn)單的通過include "asio.hpp"來調(diào)用.

#include <iostream>
#include <boost/asio.hpp>

此外,這個(gè)示例用到了timer,我們還要包含Boost.Date_Time的頭文件來控制時(shí)間.

#include <boost/date_time/posix_time/posix_time.hpp>

使用asio至少需要一個(gè)boost::asio::io_service對(duì)象.該類提供了訪問I/O的功能.我們首先在main函數(shù)中聲明它.

int main()
{
boost::asio::io_service io;

下一步我們聲明boost::asio::deadline_timer對(duì)象.這個(gè)asio的核心類提供I/O的功能(這里更確切的說是定時(shí)功能),總是把一個(gè)io_service對(duì)

象作為他的第一個(gè)構(gòu)造函數(shù),而第二個(gè)構(gòu)造函數(shù)的參數(shù)設(shè)定timer會(huì)在5秒后到時(shí)(expired).

boost::asio::deadline_timer t(io, boost::posix_time::seconds(5));

這個(gè)簡(jiǎn)單的示例中我們演示了定時(shí)器上的一個(gè)阻塞等待.就是說,調(diào)用boost::asio::deadline_timer::wait()的在創(chuàng)建后5秒內(nèi)(注意:不是等待

開始后),timer到時(shí)之前不會(huì)返回任何值.

一個(gè)deadline_timer只有兩種狀態(tài):到時(shí),未到時(shí).如果boost::asio::deadline_timer::wait()在到時(shí)的timer上調(diào)用,會(huì)立即return.

t.wait();

最后,我們輸出理所當(dāng)然的"Hello, world!"來演示timer到時(shí)了.

std::cout << "Hello, world!\n";

return 0;
}

完整的代碼:
#include <iostream>
#include <boost/asio.hpp>
#include <boost/date_time/posix_time/posix_time.hpp>

int main()
{
boost::asio::io_service io;

boost::asio::deadline_timer t(io, boost::posix_time::seconds(5));
t.wait();

std::cout << "Hello, world!\n";

return 0;
}

Boost.Asio 0.37簡(jiǎn)介(翻譯自Boost.Asio 0.37的文檔的首頁)
原文:http://asio.sourceforge.net/boost_asio_0_3_7/libs/asio/doc/

翻譯:張沈鵬 http://blog.csdn.net/zuroc or http://m.shnenglu.com/zuroc

Boost.Asio是利用當(dāng)代C++的先進(jìn)方法,跨平臺(tái),異步I/O模型的C++網(wǎng)絡(luò)庫.

入門

這個(gè)教程介紹了客戶端-服務(wù)器端的一些基本概念,同時(shí)給出了一個(gè)示例的小程序.

小程序可以啟示Boost.Asio在復(fù)雜程序中的應(yīng)用.

附注

Boost.Asio的大部分功能沒有必要編譯Boost中的其他庫,僅僅需要它們的頭文件.然而read_until和async_read_until的重載需要Boost.Regex庫(注意:MSVC 或 Borland C++的用戶需要在用戶設(shè)置中加入-DBOOST_ALL_NO_LIB來防止與Boost.Date_Time和Boost.Regex的自動(dòng)鏈接)

需要OpenSSL才可以啟用SSL支持,但Asio的其他部分不需要它.

已經(jīng)測(cè)試的平臺(tái)和編譯器:

* Win32 using Visual C++ 7.1 and Visual C++ 8.0.
* Win32 using Borland C++Builder 6 patch 4.
* Win32 using MinGW.
* Win32 using Cygwin. (__USE_W32_SOCKETS must be defined.)
* Linux (2.4 or 2.6 kernels) using g++ 3.3 or later.
* Solaris using g++ 3.3 or later.
* Mac OS X 10.4 using g++ 3.3 or later.
* QNX Neutrino 6.3 using g++ 3.3 or later.

原理:

Boost.Asio可以讓程序員用C++的程序體系取代那種需要使用system底層操作的網(wǎng)絡(luò)編程.特別的值得注意的是,Boost.Asio試圖解決一下一些問題.

*可移植性.
庫可以支持并提供一系列常用操作系統(tǒng)的一致行為.

*可測(cè)量性:
庫允許并鼓勵(lì)開發(fā)者在網(wǎng)絡(luò)編程中檢測(cè)成百或成千并發(fā)的連接數(shù).庫在各個(gè)平臺(tái)的實(shí)現(xiàn)要用這種機(jī)制來最優(yōu)的實(shí)現(xiàn)這種可測(cè)量性.

*效率:
庫要支持 分散-聚合I/O(scatter-gather I/O) 之類的技術(shù),允許協(xié)議的 最小量(minimise) 的數(shù)據(jù)交換.

*伯克利(Berkeley) sockets模型:
該模型的API被廣泛的采用和理解,被許多文獻(xiàn)介紹.其他程序語言通常使用類似網(wǎng)絡(luò)API的接口.

*易用:
降低新手使用該工具的入門障礙,勝于框架和模式.也就是說,試圖最簡(jiǎn)化前端的學(xué)習(xí),僅僅需要理解一些基本規(guī)則和指導(dǎo)方針.此外,庫的用戶僅需要理解使用到的特定函數(shù).

*可以作為進(jìn)一步抽象的基礎(chǔ):
庫應(yīng)該允許其他庫的開發(fā)者進(jìn)行更高層的抽象,比如:實(shí)現(xiàn)常用的協(xié)議Http.

盡管當(dāng)前的Boost.Asio的實(shí)現(xiàn)主要關(guān)注的是網(wǎng)絡(luò),但異步I/O可以被擴(kuò)展到其他系統(tǒng)資源,比如 文件.

由boost網(wǎng)絡(luò)庫說起...

文末這篇Email是2006-03-22的,而今天已經(jīng)2006-8-5日了,我看到asio的soureforge的主頁上說.

20 June 2006 - asio version 0.3.7 (development) released.
This release includes major interface changes arising out of the Boost review of asio.
The asio library was accepted into Boost and will appear in a future Boost release.

不過Boost release到現(xiàn)在還是

boost 1.33.1 December 5, 2005.

asio不知道還要等到什么時(shí)候才在Boost release放出,真是望穿秋水啊.
C++0x標(biāo)準(zhǔn)也是這樣,等就一個(gè)字.
由此而觀,國(guó)外C++社區(qū)真是小心謹(jǐn)慎,穩(wěn)定第一,在他們的理念中C++更類似于一件藝術(shù)品.反觀Java,.Net的日新月異,可以說是功能第一,方便第二,速度第三.商業(yè)與學(xué)院的區(qū)別真是不辨自明.

而在我的理念中,對(duì)于面向普通桌面用戶(不是專業(yè)/行業(yè)軟件)的由于功能的類似(即使你有一個(gè)很有創(chuàng)意的軟件,不出幾天就會(huì)有軟件的翻版),界面和速度可能就是生存的關(guān)鍵.對(duì)與C++的Gui這一塊,我感覺是極端不適應(yīng)(也許是我水平太差),幸而firefox和Google的一些小程序提供有了一種新的界面思路:一個(gè)后臺(tái)服務(wù)+一個(gè)微型web服務(wù)器+網(wǎng)頁界面.利用Ajax技術(shù)可以寫出很好界面來(我寫了一個(gè)默寫單詞軟件網(wǎng)頁界面的試驗(yàn)品,兼容firefox和IE6,下載地址:
http://groups.google.com/group/go2program/browse_thread/thread/5ee588253b70df77
中的附件 source.rar
)

所有在cpp@codingnow.com中高人的指點(diǎn)下我知道了asio.從而有了上面這些感慨.

下面是我翻譯的關(guān)于宣布asio成為boost一員的Email,這是我第一次逐字逐句的翻譯.水平有限(6級(jí)沒過),望多指教.

原文:Jeff Garland
翻譯:張沈鵬 http://blog.csdn.net/zuroc or http://m.shnenglu.com/zuroc

大家好-
我很高興的宣布asio(注:網(wǎng)絡(luò)庫)正式成為boost庫的一員.正如其他Boost庫的審查一樣,asio的審查產(chǎn)生了許多的討論,觀點(diǎn)和爭(zhēng)辯.評(píng)價(jià)歸結(jié)于贊美,包括在工程中成功運(yùn)用的例子,嚴(yán)肅的(庫結(jié)構(gòu))設(shè)計(jì)方面的意見.公平的說,在我看來,asio是了一個(gè)通用,可靠的庫,已經(jīng)可以被加入Boost庫--提供了開發(fā)者強(qiáng)烈要求的關(guān)鍵領(lǐng)域的應(yīng)用.

當(dāng)然,一如往常,asio還談不上完美--不少關(guān)鍵性的問題這次審評(píng)并沒有考慮.在此我僅列舉審評(píng)中許多的意見中的一部分:

- 修正動(dòng)態(tài)內(nèi)存分配的問題
- 更改接口以在運(yùn)行時(shí)ipv4和ipv6的無縫切換
- 改進(jìn)以適應(yīng)強(qiáng)類型的socket接口(????什么意思,請(qǐng)高手指教)

Chris已經(jīng)獲知了大量關(guān)于動(dòng)態(tài)內(nèi)存分的配解決方案,并且我會(huì)就接口和其他方面的問題在Boost的郵件列表上進(jìn)行跟蹤討論以期能盡善盡美.

其他可以作為進(jìn)一步增強(qiáng)重要的改進(jìn)包括:

- 盡可能減少c風(fēng)格的接口
- 和iostream進(jìn)一步的整合
- 性能優(yōu)化
- 改進(jìn)文檔(不要成為Boost w/o this one //???怎么翻譯)

Chris has a much longer list of changes garnered from the review and is well on his way to addressing many of them.
Chris有著一個(gè)更長(zhǎng)的針對(duì)審評(píng)意見的改進(jìn)計(jì)劃表,同時(shí)他與建議者有著很好的連續(xù).

注意,有幾個(gè)討論是關(guān)于性能的,這是asio所涉及領(lǐng)域的核心問題.其中一個(gè)就是上面提到的動(dòng)態(tài)內(nèi)存分配.一般而言,審評(píng)者對(duì)此的期望很高.然而,在審評(píng)討論之后,在我的印象中許多開發(fā)者發(fā)現(xiàn)asio只要改進(jìn)內(nèi)存分配,它的表現(xiàn)就已經(jīng)足夠勝任重要的工程.(但)我希望Chris在開發(fā)asio余下的時(shí)間里,可以采納審評(píng)者的一些其他方面的性能意見.

再一次我為Boost社區(qū)延遲公布審評(píng)結(jié)果而抱歉.這次推遲完全是由于我的時(shí)間安排問題,與asio無關(guān).再次感謝所有的審評(píng)者的貢獻(xiàn),尤其向Chris為Boost添加asio而作出的巨大努力致敬.

原文:
From: Jeff Garland

All -

I'm pleased to announce that asio has been accepted into Boost. As usual with a Boost review, the asio review generated plenty of discussion, issues, and controversy. Comments ranged from high praise,including success stories of projects in production, to serious design concerns and issues. On balance, in my judgment, asio provides a generally solid library that is ready for inclusion into the Boost library -- providing key functionality in an area that developers have a strong need.

Of course, like anything else, asio is not perfect -- a number of key issues were uncovered during the review. In terms of required changes I'm only going to cite a few:

- Fixes to dynamic memory allocation issues
- Interface changes to support ipv4 and ipv6 seamlessly at runtime
- Improvements to support strongly typed socket interfaces

Chris has communicated a couple possible solutions to the memory allocation issue and I'll ask that the interface and other changes for this issue continue to be discussed on the Boost list so consensus can be achieved on the best resolution.

Other key improvements that should be explored as future enhancements include:

- Possible removal of some of the c-style interfaces
- Exploration of higher level iostream integrations
- Performance improvements
- Improved documentation (wouldn't be Boost w/o this one)

Note that there were several threads and discussions about performance,which is particularly critical for the domain covered by asio. One of the performance issues is the dynamic memory allocation issue cited above. In general, the reviewers have extremely high expectations here. However, after reviewing the discussion and library it's my belief that
many developers will find asio performance sufficient to build significant projects with only the memory allocation changes. I expect Chris will be able to address some of the other performance issues cited by reviewers in asio over time.

Once again I'll apologize to the Boost community for the delay in the review results. The delay was entirely due to my own personal scheduling issues and should not reflect on asio in any way. Thanks again to all the reviewers for their effort and especially to Chris for his tremendous effort in bringing asio to Boost!

Jeff