/* This is a simple program to evaluate prefix notation
   for furture query mail contact me to yayati_2001@yahoo.com */
#include
#include 
#include
#include
int eval(char p[], int &index);

int isnum(char ch);

int isnum(char ch)
 {
  if ((ch >= '0') && (ch <= '9'))
    return 1;  // true
  else
    return 0;  // false
  }

int eval(char p[], int &index)
  {
  char ch;
  int  operand_1;
  int  operand_2;

  ch = p[index];
  index++;
  if (isnum(ch))
    return((ch-'0'));
  else
    {
    operand_1 = eval(p,index);
    operand_2 =eval(p,index);

    switch (ch)
    {
    case '+' : return operand_1 + operand_2;
    case '-' : return operand_1 - operand_2;
    case '/' : return operand_1 / operand_2;
    case '*' : return operand_1 * operand_2;
    default:
	printf("invalid do_op call " );exit(0);

    }
  }
}

void main()
  {
  clrscr();
  char exp[100];
  int  index=0;

  printf("Enter prefix expression: ");
  gets(exp);

    printf("Result is: %d n",eval(exp,index));
    getch();
 }

/*
	program trace for expression *+53-52
	result of expression is 24

	program will call eval with

			eval('*+53-52',0);
	and this is the tree showing the recursive call

				eval('*+53-52',0)
				      |
		----------------------+----------------------------
		ch='*'                                        ch ='-'
level 0       	index=1                                       index=5
		op1 = not known yet;               op2 not known yet
			|                                        |
			|                               ---------+----------
	      ----------+-----------                  ch=5             ch=2
level 1	      ch='+'  op1=8                           index=6          index=7
	      index=2                             (here ch is num   (and op2=2
		|                                  hence op1 =5)
		|
level 2	   -----+------
	   |          |
	   |	      |
	ch=5;        ch=3  (here ch is num hence
	index=3      index=4 returns 5 as op1 and
			     3 as op2)

The problems are entailed below:

1. The algorithm doesn't handle power (Math.pow(a,b)) expression.
which is of higher priority when evaluating the prefix expression.
2. The binary tree algorithm can handle only 2 jointed prefix expression,
because the tree is NOT balanced. i.e.
it cannot handle the following expression of 3 expresion component :
13 + 0.6 / ( -3 + 1.75 ) * 2 ^ 3 – 2.35 ==>
- + 13 * / 0.6 + -3 1.75 ^ 2 3 2.35
3. The infix parser is NOT formal
Strictly speaking, those operators with lowest priority come first.

Hence, the algorithm is NOT 100% true.

To generate prefix/postfix/infix expression from one of else,
we need to build up an expression tree ONCE,
then perform prefix/postfix/infix transversal to directly evaluate it.

I have found one of the useful java code to build such a tree.

/**
 * Expression Tree of TreeNode objects, built from a prefix expression.
 *
 * NOTE:  Specific to Java version 5.0 --- Scanner
 *
 * The nodes are built by the recursive method "build", which calls
 * itself for internal nodes; e.g.: node.setRight ( build ( input ) );
 *
 * Beyond construction, this supports display as prefix expression,
 * postfix expression, and as parenthesized infix expression, as well
 * as evaluation of the expression, returning the value;
 *
 * @author  Timothy Rolfe
 */
import java.util.Scanner;    // Specific to Java 1.5.x

public class ExpressionTree
{
   /**
    * One node in an expression tree, allowing double values.
    *
    * @author  Timothy Rolfe
    */
   private static class TreeNode
   {
      private final boolean  leaf;   // ?Is this a leaf? else internal
      private final char     op;     // For an internal node, the operator
      private       double   value;  // For a leaf, the value
      private       TreeNode left,   // Left subexpression for an internal node
                             right;  // Right subexpression

      // Bare-bones constructor
      private TreeNode ( boolean leaf, char op, double value )
      {
         this.leaf    = leaf;
         this.op      = op;
         this.value   = value;
         this.left    = null;   // Empty to start
         this.right   = null;
      }

      // For leaf nodes, show the value; for internal, the operator.
      public String toString()    // To override Object.toString, must be public.
      {  return leaf ? Double.toString(value) : Character.toString(op) ;  }
   }

   TreeNode root = null;

   public ExpressionTree ( Scanner input )
   {  root = build(input);  }

/**
 * Based on a white-space delimited prefix expression, build the
 * corresponding binary expression tree.
 * @param  input  The scanner with the expression
 * @return reference to the corresponding binary expression tree
 */
   private TreeNode build ( Scanner input )
   {
      boolean  leaf;
      String   token;
      double   value;
      TreeNode node;

      leaf = input.hasNextDouble();
      if ( leaf )
      {
         value = input.nextDouble();
         node = new TreeNode ( leaf, '', value );
      }
      else
      {
         token = input.next();
         node  = new TreeNode ( leaf, token.charAt(0), 0.0 );
         node.left  = build ( input );
         node.right = build ( input );
      }
      return node;
   }

/**
 * Show the expression tree as a postfix expression.
 * All the work is done in the private recursive method.
 */
   public void showPostFix ()
   {
      showPostFix ( root );
      System.out.println();
   }

   // Postfix expression is the result of a post-order traversal
   private void showPostFix ( TreeNode node )
   {
      if ( node != null )
      {
         showPostFix ( node.left );
         showPostFix ( node.right );
         System.out.print ( node + " " );
      }
   }

/**
 * Show the expression tree as a prefix expression.
 * All the work is done in the private recursive method.
 */
   public void showPreFix ()
   {
      showPreFix ( root );
      System.out.println();
   }

   // Prefix expression is the result of a pre-order traversal
   private void showPreFix ( TreeNode node )
   {
      if ( node != null )
      {
         System.out.print ( node + " " );
         showPreFix ( node.left );
         showPreFix ( node.right );
      }
   }

/**
 * Show the expression tree as a parenthesized infix expression.
 * All the work is done in the private recursive method.
 */
   public void showInFix ()
   {
      showInFix ( root );
      System.out.println();
   }

   // Parenthesized infix requires parentheses in both the
   // pre-order and post-order positions, plus the node
   // itself in the in-order position.
   private void showInFix ( TreeNode node )
   {
      if ( node != null )
      {
         // Note:  do NOT parenthesize leaf nodes
         if ( ! node.leaf )
            System.out.print ("( ");        // Pre-order position
         showInFix ( node.left );
         System.out.print ( node + " " );   // In-order position
         showInFix ( node.right );
         if ( ! node.leaf )                 // Post-order position
            System.out.print (") ");
      }
   }

/**
 * Evaluate the expression and return its value.
 * All the work is done in the private recursive method.
 * @return  the value of the expression tree.
 */
   public double evaluate ()
   {  return root == null ? 0.0 : evaluate ( root ) ;  }

   // Evaluate the expression:  for internal nodes, this amounts
   // to a post-order traversal, in which the processing is doing
   // the actual arithmetic.  For leaf nodes, it is simply the
   // value of the node.
   private double evaluate ( TreeNode node )
   {
      double result;    // Value to be returned

      if ( node.leaf )        // Just get the value of the leaf
         result = node.value;
      else
      {
         // We've got work to do, evaluating the expression
         double left, right;
         char   operator = node.op;

         // Capture the values of the left and right subexpressions
         left  = evaluate ( node.left );
         right = evaluate ( node.right );

         // Do the arithmetic, based on the operator
         switch ( operator )
         {
            case '-':  result = left - right;  break;
            case '*':  result = left * right;  break;
            case '/':  result = left / right;  break;
            case '^':  result = Math.pow (left, right );  break;
         // NOTE:  allow fall-through from default to case '+'
            default:   System.out.println ("Unrecognized operator " +
                          operator + " treated as +.");
            case '+':  result = left + right;  break;
         }
      }
      // Return either the leaf's value or the one we just calculated.
      return result;
   }
}

Then to evaluate it, we need to compute an main(args[]) program.

/**
 * Prefix calculator:  generate the expression tree, then display it
 * in the various supported means and finally show the result of the
 * calculation.
 *
 * NOTE:  Specific to Java version 5.0 --- Scanner
 *
 * @author  Timothy Rolfe
 */
import java.util.Scanner;

public class PrefixCalc
{
   public static void main ( String[] args )
   {
      ExpressionTree calc;

   // Allow for a command-line argument (which would be double-quoted).
      if ( args.length > 0 )
      {
         System.out.println ("Processing string " + args[0]);

         calc = new ExpressionTree(new Scanner(args[0]));
      }
      else
      {
         System.out.println
           ( "Prefix expression, with all elements separated by blanks");

         calc = new ExpressionTree(new Scanner(System.in));
      }

      System.out.println ("\nInput as prefix expression:");
      calc.showPreFix();

      System.out.println ("\nInput as postfix expression:");
      calc.showPostFix();

      System.out.println ("\nInput as parenthesized infix expression:");
      calc.showInFix();

      System.out.println ("\nValue:  " + calc.evaluate());
   }
}

I hope this will help for newbie of Object-Orientated Java programmers.

In order to make PC able to read the flash drive contents in the mobile phone,
we need the concerned USB Flash Driver:

http://sites.google.com/site/aikosunoo/Home/USBFlash_driver.rar?attredirects=0

The follow the steps:

0. unpack the downloaded archive to a directory.
1. Switch OFF your phone !
2. Remove the battery !
3. Reinsert your battery !
4. Don’t start your phone !
5. HOLDING ‘C’ on your phone, connect your phone to USB Cable !
6. Don’t Leave C through out driver installation.

Note: If step 5 and 6 failed, try the buttons '2' + '5' instead.

You get a notification in System Tray ! Saying SEMC USB Flash Device

A new window should appear “Found new hardware wizard”
It asks permissions for to search for drivers for the device.
Select “No, not this time” and click next.

Then select the option
"INSTALL FROM A SPECIFIED LOCATION(ADVANCED)" and click next

Then choose “Search for the best drivers in these locations”,
and check the “Include this location in this search”,
now browse for the downloaded drivers.

Click next, Wait for some time, drivers get installed.

Now…….

Open the directory, where you unpacked the archive, I
n the directory, you will find a ggsemc.inf file.
Click on it with the right mouse button, and then click ‘Install‘.

Video tutorial: http://www.4shared.com/file/43289822/36c36144/usbflash.html

Microsoft Windows XP do have such a powertoy:

http://download.microsoft.com/download/whistler/Install/2/WXP/EN-US/CmdHerePowertoySetup.exe

Then you can execute your javac/appletviewer program at any rate. 

In order to make words as artifact,
you must need to know how to edit the shape of fonts/types.

Remember, there is no freelunch.
i.e. There exists NO software which can generate the logo you want once-and-for-all.
You will have to pay by editing every anchors of the shattered fonts/types as curves.

So, You can try using Adobe Illustrator to mimic the Japanese term  of
ojamajo, おジャ魔女どれみ:

as

Wrong choice of typefaces,
some of which you may need to buy in case of Dynalab (華康),
will lead to fatal flaw of your artwork. 

Yes, design - kinda art - is right oja.

http://rapidshare.com

It only allow 10-time download per IP and 90 days for store.
What's more, you have to wait for at least 15 minute before next download!

Filefactory also sucks!

http://filefactory.com/

High upload speed huh?

If you upload more than 10MB, the uploader halts!

So, the best choice is http://sendspace.com/

You can upload a 200-MB file within 3 minute,
and download it at at least 150MB/s.

Using simply embed tag and insert  parameters:

autoplay="false"
src="[destination]" //(say: try.amr)

Then in Firefox the amr file can be played.
Note that in IE7, ActiveX warning message may appear. Simply enable it.

Then I use a Philips mic from a broken headphone to record it.
All seems work fine, but the filesize is too big!
(the mp3 file streamed out using Audacity is around 1MB per minute).

I have once thought of using xanga audio function to proceed such podcast,
but the upload and transcoding speed is of tortoise level, so I gave it up.

Next, I have to chop down the filesize using streaming technology.
I choose Windows Media Video instead of RMVB,
because wmv is the default format using Windows Movie Maker.
It is one of the podcasting solution in conjunction with youtube and webcam.

But recently I wanna make the venue of podcasting flexible,
and I discovered my Sony Erricson K800i phone has recording function
inside "Entertainment" module.  A fucking annoying point is that,
the "pause" function is not provided, although I can really record
everywhere and with 1kb per second.
The fileformat is Adaptive Multi-Rate (AMR).
Then before I xanga-audio it I need some amr2mp3-like softwares.

The required freeware can be found here:

http://aikosunoo.googlepages.com/amr2mp3.cab

OJA STUDIES - Textbook and/or Related Integral Social Science Discourses (龜相學 - 跨科文獻)

許多人問過我:「你是如何用 Macromedia FlashPaper 2 製造試卷?」

沒有錯, 我用Macromedia FlashPaper 2已經超過30日,
許多人以為我用盜版, 其實它在我的電腦仍是試用版。

只要你有 reborn card (using PCI slot),你就可以不用洗機地無限期試用這那軟件:

1. 用 reborn card backup 整個harddisk,
期間不要安裝任何不需 reboot computer 的 trial shareware
2. 設定reborn card 為「每次 boot 機時回復/重置所有 harddisk 資料」
3. 利用 reborn card 開機後, 所有附加 Windows Registry 記錄就會在關機後被刪
4. 開機後類似 Flashpaper 的軟件才可以安裝(每次開機必須重新安裝)

當然, 如果個 software 用 network license, 上述方法等同廢話...

import java.io.*;

public class runDOS {
 public static void main(String[] args) throws IOException {
  System.out.println("Hello Word");
  String[] command =  new String[4];
           command[0] = "cmd";
           command[1] = "/C";
           command[2] = "dir";
           command[3] = "c:＼＼";
  Process p = Runtime.getRuntime().exec(command);
          BufferedReader stdInput = new BufferedReader(new InputStreamReader(p.getInputStream()));
  BufferedReader stdError = new BufferedReader(new InputStreamReader(p.getErrorStream()));

  // read the output from the command

          String s = null;
           System.out.println("Here is the standard output of the command:\n");
           while ((s = stdInput.readLine()) != null) {
               System.out.println(s);
          }
  // read any errors from the attempted command

           System.out.println("Here is the standard error of the command (if any):\n");
           while ((s = stdError.readLine()) != null) {
                System.out.println(s);
           }        
  System.out.println("Bye!");
 }
}

Console Result:

Hello Word
Here is the standard output of the command:

 磁碟區 C 中的磁碟沒有標籤。
 磁碟區序號:  443E-6C01

 c:＼ 的目錄

2007/08/06  下午 06:48                 0 AUTOEXEC.BAT
2007/08/06  下午 06:48                 0 CONFIG.SYS
2008/08/17  下午 06:38    <DIR>          Documents and Settings
2008/07/23  下午 09:52    <DIR>          Downloads
2008/03/29  上午 03:29    <DIR>          FlexLM
2008/01/27  下午 08:59           413,696 FLVSplitter.ax
2008/03/28  下午 02:40    <DIR>          IDE
2008/05/03  上午 04:29           230,424 img2-001.raw
2008/05/03  下午 10:50           230,424 img2-002.raw
2008/03/27  下午 07:34    <DIR>          Inetpub
2004/01/08  上午 11:38           208,896 lame_enc.dll
2002/01/05  上午 03:38            54,784 msvci70.dll
2008/12/05  下午 04:40    <DIR>          Program Files
2002/08/22  下午 01:50             8,514 readme.htm
2002/08/22  下午 01:51             3,248 readme.txt
2008/03/27  下午 07:10    <DIR>          speakd03
2002/02/13  下午 01:35             3,280 web.config
2008/12/05  下午 04:42    <DIR>          WINDOWS
2008/03/27  下午 06:56    <DIR>          WINNT
2008/03/27  下午 11:23    <DIR>          WTK2.5.2
2008/03/27  下午 08:55    <DIR>          XPPEN
              10 個檔案       1,153,266 位元組
              11 個目錄  95,636,680,704 位元組可用
Here is the standard error of the command (if any):

Bye!

遞迴方法 (recursive function)

當大家學完method之後有無諗過，一個方法可唔可以呼叫返方法自身LE?!

事實上係Java當中係可以GE，而我地就會稱E類演算法為遞迴。
雖然遞迴方法通可以被for, while, do....while等迴圈所取代，而且迴圈方法更加有效率~
但係係邏輯性、可讀性、彈性方面，遞迴就會比迴圈好好多LA~

E+先比一個經典例子大家LA ~~~ 就係階乘 ( 6 ! = 6 * 5 * 4 * 3 * 2 * 1 )

public class app{
    public static void main(String[] args) {
        System.out.println("6! = " + f(6));
    }
    public static int f(int a) {
        if (a == 1)
            return 1;
        else
            return a * f(a - 1);
    }
}

First of all,
in order to unveil his hypocrisy and his malpractical psyche,
let us examine what tricks Ben Ng has used to play himself up:

Web2.0是否「美麗新世界」？日人的觀點

http://www.cuhkacs.org/~benng/Bo-Blog/read.php?973#topreply
(DANGER: DO NOT LINK TO THIS SITE DIRECTLY!
EITHER USE PROXY SERVERS,
OR COPY-AND PASTE THE URL AFTER CLOSING ALL BROWSERS!)

Web2.0是第二代互聯網之意
（相對於1990年代單向資訊發佈為主的Web1.0），
從千禧年至今都是Web2.0的時代，
Web成為人類最重要的資訊、商業與生活平台，
分享與互動是Web2.0時代的核心精神， Blog、Wiki、
Goggle、Facebook、Youtube、Flickr、
iPod都是Web2.0的產品。對於Web2.0時代的來臨，
有人快樂有人愁。
樂觀者歡呼它為一般平民百姓帶來前所未有的自由、
平等、民主與互動精神；
悲觀者擔心當權者利用法律及監察系統控制互聯網、
人性的醜惡阻礙真正的分享與互動及知識庸俗化等問題。
這種對Web2.0革命的兩極觀點在日本也同樣掀起爭論。

樂觀派的代表是IT界名人梅田望夫，
他寫了多本有關網絡社會的書，
其中以《ウェブ進化論》（Web進化論，2006, 筑摩）
為代表。書中強調Web2.0的精神
「不是被動的享受服務，而是主動、積極的參與其中」。
梅田表示在Web2.0的時代，
知識及資訊不再是某階層所獨有，只要在網上尋找，
什麼也會出來。發言權也不再是政府及傳媒所獨享，
創作及做生意的大門也為一般人大開。
Web2.0將帶來財富及社會地位的「大執位」，
只要有腦，白手也可興家。
他勸喻每個人應積極在網絡社會尋找新的個人定位與機會。

悲觀派代表是有「情報學第一人」之稱的東京大學教授
西垣通。他在《ウェブ社会をどう生きるか》
（如何在網絡社會存活下去？2007，岩波）
對樂觀派展開嚴厲批評，
指出他們多是為了個人私利而唱好Web2.0，
而忽略它帶來的負面影響。首先，年長一代被邊緣化，
(Ben Ng is acting valetudinarianism here)
他們很多連上網都不會，
愈來愈多的網上登記及網上服務令他們覺得被疏離。
Web2.0帶來的不是自由平等社會，
而且將「差格社會」強化。第二，西垣批評情報泛濫，
但質素卻不斷下降，不可靠情報充斥，
(Ben Ng is arrogant enough to think of his blog as reliable) 
而且互相抄襲情況嚴重，對知識傳授並無好處。
第三，至於人人均可為作家、記者之平民機會主義，
西垣認為是一廂情願的想法，
因為事實上網絡世界根本就是「差格社會」
及「弱肉強食的世界」，著名的網站及博客愈強，
(He is quite ignorant enough to think that he is strong)
無名網站及博客愈弱。
(He thinks that my blog lack fame, which is an imbecile deed)

以上兩派均有其道理，
(Because he is acting defense mechanism of rationalization)
代表Web2.0好壞並存的兩面。
作為生活在Web2.0時代的人，
我們一方面要如梅田望夫所言，
應積極在網絡社會尋找新的個人定位與機會。
(He incites audiences to be opportunitists 機會主義者/
speculative 投機的)
另一方面，西垣通指出的問題也十分真實，
(That means all other bloggers are uttering fictional bullshit?)
所以要鍛練自己，
(He means that he wanna upgrade his own IT knowledge,
with the fact that his is an IT idiot.
In truth, it is questionable if he wanna be hacker.)
把握如何在網上選取可靠的資訊、
(He brags his blog as only reliable one among all ACG blogs)
參與高質的網上社區
(He also brags his blog as high-quality arrogantly,
which is to be disproven here. Also, he implies that
his enemies' arena, such as HKGolden and mysinablog,
are of lower quality than his blog.)
及提升個人的網上技術
(He means that he will trace your browsing record,
such as referrer and IP, and transit such information
to his friends as well as hackers)
與品格
(He alleges that, unlike HKGolden-ers,
he never say foul language,
and he imbecilily believe that this alone can,
to him, prove his righteousness which is pseudo-true,
and is also to be disproven here!)
也十分重要。

首先, 我要說明為何我要屌鳩吳偉明,
因為他不是正人君子, 懶正義, 懶偉大, 大男人得很行為亦像港女,玩暗示。
看看他的友好連結後,
我相信任何女性或女性主義者(包括御宅族)也會不屑看他的偽善blog!

他調查別人身份的技術是非常陽春的說......

http://www.cuhkacs.org/counter/index.php?l=&name=benng2&action=referers&page=0

留意他是如何Record 別人的IP 和 Referrer,
你就可以知道大量無知、為乞求身份認同而盲目支持他的ACG迷同佢一樣咁幼稚...

批判他的blog/forum url (包括 HKGolden, 本網)永遠列為blockedReferrer,
這樣一來他就可以成為 spin-doctor, 虛偽地隱惡揚善,
實際上他逃避現實, 不肯反省為何有人憎恨他及他的言行(妄自菲薄, 鳩up , 莫須有),
繼續為面子死撐, 連敵人的心理也不肯去面對, 理解,
仍認為自己識小小日本文化和自己偏私的價值觀為absolutely correct「真理」,
扮代表, 企圖製造知日霸權(hegemony), 壟斷流行文化言論空間。
真正的文化高手(e.g. 預科真光中學教師)肯定會當他是Black Sheep。
因此, 吳偉明挪用小道日本文化資訊聊以自慰, 乃不智之舉!

知日為智? 弱智!