c++

Libcurl 简明使用指南

Michael — Tue, 29 Nov 2005 09:03:00 GMT

Libcurl为一个免费开源的，客户端url传输库，支持FTP，FTPS，TFTP，HTTP，HTTPS，GOPHER，TELNET，DICT，FILE和LDAP，跨平台，支持Windows，Unix，Linux等，线程安全，支持Ipv6。并且易于使用。

http://curl.haxx.se/libcurl/

从http://curl.haxx.se/libcurl/ 下载一个稳定的版本，注意选择OS。

编译libcurl

下载下来的是源码包，需要编译。

解压zip文件，进入curl-7.14.0\lib目录（我下载的是7.14.0）。

编译Debug版本。新建一个批处理bat文件，如buildDebug.bat，内容如下：

call "C:\Program Files\Microsoft Visual Studio\VC98\Bin\vcvars32.bat"

set CFG=debug-dll-ssl-dll-zlib-dll

set OPENSSL_PATH=E:\SSL\openssl-0.9.7e

set ZLIB_PATH=E:\zip\zlib123

nmake -f Makefile.vc6

其输出：libcurld_imp.lib, libcurld.dll

编译Release版本。新建一个批处理文件BuildRelease.bat，内容如下：

call "C:\Program Files\Microsoft Visual Studio\VC98\Bin\vcvars32.bat"

set CFG=release-dll-ssl-dll-zlib-dll

set OPENSSL_PATH=E:\SSL\openssl-0.9.7e

set ZLIB_PATH=E:\zip\zlib123

nmake -f Makefile.vc6

其输出：libcurl_imp.lib, libcurl.dll

上面编译的是libcurl的 dll，使用OpenSSL Dll版本和Zlib Dll版本。如果没有，可以从www.openssl.org 或者http://www.zlib.net/ 下载。

如果需要编译其他版本，可查看Makefile.vc6，设定相应的CFG 参数即可。

商业软件使用libcurl时，只需要包含其copywrite声明即可。

Sample

#include <stdio.h>
#include "../curl-7.14.0/include/curl/curl.h"
#pragma comment(lib, "../curl-7.14.0/lib/libcurl_imp.lib")

int main(void)
{
  curl = curl_easy_init();
  if(curl) {

    CURLcode res;
    res = curl_easy_setopt(curl, CURLOPT_PROXY, "Test-pxy08:8080");
    res = curl_easy_setopt(curl, CURLOPT_PROXYTYPE, CURLPROXY_HTTP);
    res = curl_easy_setopt(curl, CURLOPT_URL, "http://www.vckbase.com");
    res = curl_easy_perform(curl);

    if(CURLE_OK == res) {
      char *ct;
      /* ask for the content-type */
      /* http://curl.haxx.se/libcurl/c/curl_easy_getinfo.html */
      res = curl_easy_getinfo(curl, CURLINFO_CONTENT_TYPE, &ct);

      if((CURLE_OK == res) && ct)
        printf("We received Content-Type: %s ", ct);
    }

    /* always cleanup */
    curl_easy_cleanup(curl);
  }
  return 0;
}

已知bugs

不支持SOCKS4。

SOCKS5实现有缺陷，导致必须设置用户和密码。

连接proxy过程不能设置超时。

Google™ Code Jam - 中国编程挑战赛模拟题三

Michael — Sat, 26 Nov 2005 14:20:00 GMT

Google™ Code Jam - 中国编程挑战赛

Problem Statement

When editing a single line of text, there are four keys that can be used to move the cursor: end, home, left-arrow and right-arrow. As you would expect, left-arrow and right-arrow move the cursor one character left or one character right, unless the cursor is at the beginning of the line or the end of the line, respectively, in which case the keystrokes do nothing (the cursor does not wrap to the previous or next line). The home key moves the cursor to the beginning of the line, and the end key moves the cursor to the end of the line.

You will be given a int, N, representing the number of character in a line of text. The cursor is always between two adjacent characters, at the beginning of the line, or at the end of the line. It starts before the first character, at position 0. The position after the last character on the line is position N. You should simulate a series of keystrokes and return the final position of the cursor. You will be given a string where characters of the string represent the keystrokes made, in order. 'L' and 'R' represent left and right, while 'H' and 'E' represent home and end.

Definition

Class:	CursorPosition
Method:	getPosition
Parameters:	string, int
Returns:	int
Method signature:	int getPosition(string keystrokes, int N)
(be sure your method is public)

Constraints

keystrokes will be contain between 1 and 50 'L', 'R', 'H', and 'E' characters, inclusive.

N will be between 1 and 100, inclusive.

Examples

"ERLLL"

Returns: 7

First, we go to the end of the line at position 10. Then, the right-arrow does nothing because we are already at the end of the line. Finally, three left-arrows brings us to position 7.

"EHHEEHLLLLRRRRRR"

Returns: 2

All the right-arrows at the end ensure that we end up at the end of the line.

"ELLLELLRRRRLRLRLLLRLLLRLLLLRLLRRRL"

Returns: 3

"RRLEERLLLLRLLRLRRRLRLRLRLRLLLLL"

Returns: 12

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

我的解答

#include <string>

using namespace std;

const char LEFT = 'L';
const char RIGHT = 'R';
const char HOME = 'H';
const char END = 'E';

class CursorPosition
{
public:
    int getPosition(string keystrokes, int N);
};

int CursorPosition::getPosition(string keystrokes, int N)
{
    int nCurrPos = 0;

    int i=0;
    for (i = 0; i < keystrokes.length(); i++)
    {
        if (LEFT == keystrokes.at(i))
        {
            if (nCurrPos > 0)
            {
                nCurrPos--;
            }
        }
        else if (RIGHT == keystrokes.at(i))
        {
            if (nCurrPos < N)
            {
                nCurrPos++;
            }
        }
        else if (HOME == keystrokes.at(i))
        {
            nCurrPos = 0;
        }
        else if (END == keystrokes.at(i))
        {
            nCurrPos = N;
        }
    }

    return nCurrPos;
}

Google™ Code Jam - 中国编程挑战赛模拟题二

Michael — Sat, 26 Nov 2005 14:18:00 GMT

Google™ Code Jam - 中国编程挑战赛 http://www.topcoder.com/gcjc_zh

题目如下：

Problem Statement

A square matrix is a grid of NxN numbers. For example, the following is a 3x3 matrix:

 4 3 5
 2 4 5
 0 1 9

One way to represent a matrix of numbers, each of which is between 0 and 9 inclusive, is as a row-major string. To generate the string, simply concatenate all of the elements from the first row followed by the second row and so on, without any spaces. For example, the above matrix would be represented as "435245019".

You will be given a square matrix as a row-major string. Your task is to convert it into a vector <string>, where each element represents one row of the original matrix. Element i of the vector <string> represents row i of the matrix. You should not include any spaces in your return. Hence, for the above string, you would return {"435","245","019"}. If the input does not represent a square matrix because the number of characters is not a perfect square, return an empty vector <string>, {}.

Definition

Class:	MatrixTool
Method:	convert
Parameters:	string
Returns:	vector <string>
Method signature:	vector <string> convert(string s)
(be sure your method is public)

Constraints

s will contain between 1 and 50 digits, inclusive.

Examples

"435245019"

Returns: {"435", "245", "019" }

The example above.

"9"

Returns: {"9" }

"0123456789"

Returns: { }

This input has 10 digits, and 10 is not a perfect square.

"3357002966366183191503444273807479559869883303524"

Returns: {"3357002", "9663661", "8319150", "3444273", "8074795", "5986988", "3303524" }

#include <string>
#include <vector>
#include <math.h>

using namespace std;

class MatrixTool
{
public:
    vector<string> convert(string s);
};

vector<string> MatrixTool::convert(string s)
{
    vector<string> vecResult;

    int iLength = s.length();
    int iYinzi = (int)sqrt(iLength);
    if (iYinzi * iYinzi != iLength)
    {
        return vecResult;
    }

    int i = 0;
    for (; i < iYinzi; i++)
    {
        string strTmp = s.substr(iYinzi*i, iYinzi);
        vecResult.push_back(strTmp);
    }

    return vecResult;
}

模拟题比较简单。

Google™ Code Jam - 中国编程挑战赛模拟题一

Michael — Sat, 26 Nov 2005 14:15:00 GMT

Google™ Code Jam - 中国编程挑战赛
http://www.topcoder.com/gcjc_zh

你有用技术改变世界的梦想吗？你有挑战难度的决心吗？你想和国内计算机精英一决高下吗？全球编程界知名的“Google编程挑战赛Code Jam”即将登陆中国。这项比赛每年都是全球计算机界的一次盛事。今年 Google 首次专门为中国举办这项比赛，旨在弘扬计算机科学的艺术，推进中国计算机编程教育，鼓励并嘉奖中国顶级编程人才。

竞赛的题目具有相当的挑战性，竞赛奖品也非常丰厚。有志之士可借此机会一展才能，成为脱颖而出的中国最佳。

这里有极富挑战性的题目，高科技的奖品，以及令人赞叹的荣耀，你还在等什么？

趁周末，申请一个帐号，进入Try了一下模拟题。
题目是英文描述：

Problem Statement

A simple line drawing program uses a blank 20 x 20 pixel canvas and a directional cursor that starts at the upper left corner pointing straight down. The upper left corner of the canvas is at (0, 0) and the lower right corner is at (19, 19). You are given a vector <string>, commands, each element of which contains one of two possible commands. A command of the form "FORWARD x" means that the cursor should move forward by x pixels. Each pixel on its path, including the start and end points, is painted black. The only other command is "LEFT", which means that the cursor should change its direction by 90 degrees counterclockwise. So, if the cursor is initially pointing straight down and it receives a single "LEFT" command, it will end up pointing straight to the right. Execute all the commands in order and return the resulting 20 x 20 pixel canvas as a vector <string> where character j of element i represents the pixel at (i, j). Black pixels should be represented as uppercase 'X' characters and blank pixels should be represented as '.' characters.

Definition

Class:	DrawLines
Method:	execute
Parameters:	vector <string>
Returns:	vector <string>
Method signature:	vector <string> execute(vector <string> commands)
(be sure your method is public)

Notes

The cursor only paints the canvas if it moves (see example 1).

Constraints

commands will contain between 1 and 50 elements, inclusive.

Each element of commands will be formatted as either "LEFT" or "FORWARD x" (quotes for clarity only), where x is an integer between 1 and 19, inclusive, with no extra leading zeros.

When executing the commands in order, the cursor will never leave the 20 x 20 pixel canvas.

Examples

{"FORWARD 19", "LEFT", "FORWARD 19", "LEFT", "FORWARD 19", "LEFT", "FORWARD 19"}

Returns: 
{"XXXXXXXXXXXXXXXXXXXX",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "X..................X",
 "XXXXXXXXXXXXXXXXXXXX" }

This sequence of commands draws a 20 x 20 outline of a square. The cursor is initially at (0, 0) pointing straight down. It then travels to (0, 19) after the first FORWARD command, painting each pixel along its path with a '*'. It then rotates 90 degrees left, travels to (19, 19), rotates 90 degrees left, travels to (19, 0), rotates 90 degrees left, and finally travels back to (0, 0).

{"LEFT", "LEFT", "LEFT", "LEFT", "LEFT", "LEFT", "LEFT", "LEFT"}

Returns: 
{"....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "...................." }

The cursor spins round and round, but never actually paints any pixels. The result is an empty canvas.

{"FORWARD 1"}

Returns: 
{"X...................",
 "X...................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "....................",
 "...................." }

Going forward by one pixel creates a line that is 2 pixels long because both the start and end points are painted.

{"LEFT", "FORWARD 19", "LEFT", "LEFT", "LEFT",
 "FORWARD 18", "LEFT", "LEFT", "LEFT", "FORWARD 17",
 "LEFT", "LEFT", "LEFT", "FORWARD 16", "LEFT",
 "LEFT", "LEFT", "FORWARD 15", "LEFT", "LEFT", "LEFT",
 "FORWARD 14", "LEFT", "LEFT", "LEFT", "FORWARD 13",
 "LEFT", "LEFT", "LEFT", "FORWARD 12", "LEFT", "LEFT",
 "LEFT", "FORWARD 11", "LEFT", "LEFT", "LEFT", "FORWARD 10",
 "LEFT", "LEFT", "LEFT", "FORWARD 9", "LEFT", "LEFT",
 "LEFT", "FORWARD 8", "LEFT", "LEFT", "LEFT", "FORWARD 7"}

Returns: 
{"XXXXXXXXXXXXXXXXXXXX",
 "...................X",
 "..XXXXXXXXXXXXXXXX.X",
 "..X..............X.X",
 "..X.XXXXXXXXXXXX.X.X",
 "..X.X..........X.X.X",
 "..X.X.XXXXXXXX.X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.X........X.X.X",
 "..X.X.XXXXXXXXXX.X.X",
 "..X.X............X.X",
 "..X.XXXXXXXXXXXXXX.X",
 "..X................X",
 "..XXXXXXXXXXXXXXXXXX",
 "...................." }

#include <string>
#include <vector>

using namespace std;

const char* const pcstrBlankLine = "..";

const char* const CMD_FORWARD = "FORWARD";
const int SIZE_FORWARD = strlen(CMD_FORWARD);
const char* const CMD_LEFT = "LEFT";
const int SIZE_LEFT = strlen(CMD_LEFT);

const char CANVAS_BLACK = 'X';

class DrawLines
{
public:
    enum
    {
        TYPE_NONE = 0,
        TYPE_FORWARD = 1,
        TYPE_LEFT = 2,

        CANVAS_WIDTH = 20,
        CANVAS_HEIGHT = 20,

        DIRECT_DOWN = 0,
        DIRECT_LEFT = 1,
        DIRECT_UP = 2,
        DIRECT_RIGHT = 3
    };

    vector<string> execute(vector<string> commands);

    void ExcuteOneCommand(const string& cmd);
    int GetComType(const string& strCmd, int& iStep);

    DrawLines() : m_iDirection(DIRECT_DOWN),
        m_iCurrHorizonPos(0),
        m_iCurrVerticalPos(0)
    {
    }

public:
    vector<string> m_paintedCanvas;

    int m_iDirection;
    int m_iCurrHorizonPos;
    int m_iCurrVerticalPos;
};

vector<string> DrawLines::execute(vector<string> commands)
{
    m_paintedCanvas.clear();
    int i = 0;
    for (i = 0; i < CANVAS_HEIGHT; i++)
    {
        m_paintedCanvas.push_back(pcstrBlankLine);
    }

    for (i = 0; i < commands.size(); i++)
    {
        ExcuteOneCommand(commands[i]);
    }

    return m_paintedCanvas;
}

void DrawLines::ExcuteOneCommand(const string& cmd)
{
    int iStep = 0;
    if (GetComType(cmd, iStep) == TYPE_FORWARD)
    {
        if (iStep <= 0)
        {
            return;
        }

        //Drawline
        if (DIRECT_DOWN == m_iDirection)
        {
            int nReachPos = (m_iCurrVerticalPos+iStep < CANVAS_HEIGHT)?(m_iCurrVerticalPos+iStep):CANVAS_HEIGHT-1;
            for (int i = m_iCurrVerticalPos; i <= nReachPos; i++)
            {
                m_paintedCanvas[i].at(m_iCurrHorizonPos) = CANVAS_BLACK;
            }

            m_iCurrVerticalPos = nReachPos;
        }
        else if (DIRECT_UP == m_iDirection)
        {
            int nReachPos = (m_iCurrVerticalPos-iStep >= 0)?(m_iCurrVerticalPos-iStep):0;
            for (int i = m_iCurrVerticalPos; i >= nReachPos; i--)
            {
                m_paintedCanvas[i].at(m_iCurrHorizonPos) = CANVAS_BLACK;
            }

            m_iCurrVerticalPos = nReachPos;
        }
        else if (DIRECT_LEFT == m_iDirection)
        {
            int nReachPos = (m_iCurrHorizonPos+iStep < CANVAS_HEIGHT)?(m_iCurrHorizonPos+iStep):CANVAS_HEIGHT-1;
            for (int i = m_iCurrHorizonPos; i <= nReachPos; i++)
            {
                m_paintedCanvas[m_iCurrVerticalPos].at(i) = CANVAS_BLACK;
            }

            m_iCurrHorizonPos = nReachPos;
        }
        else if (DIRECT_RIGHT == m_iDirection)
        {
            int nReachPos = (m_iCurrHorizonPos-iStep >= 0)?(m_iCurrHorizonPos-iStep):0;
            for (int i = m_iCurrHorizonPos; i >= nReachPos; i--)
            {
                m_paintedCanvas[m_iCurrVerticalPos].at(i) = CANVAS_BLACK;
            }

            m_iCurrHorizonPos = nReachPos;
        }
    }
    else if (GetComType(cmd, iStep) == TYPE_LEFT)
    {
        if (DIRECT_RIGHT == m_iDirection)
        {
            m_iDirection = DIRECT_DOWN;
        }
        else
        {
            m_iDirection++;
        }
    }
}

int DrawLines::GetComType(const string& strCmd, int& iStep)
{
    const char* pstrCmd = strCmd.c_str();
    if (strncmp(pstrCmd, CMD_FORWARD, SIZE_FORWARD) == 0)
    {
        pstrCmd += SIZE_FORWARD + 1;
        iStep = atoi(pstrCmd);
        return TYPE_FORWARD;
    }
    else if (strncmp(pstrCmd, CMD_LEFT, SIZE_LEFT) == 0)
    {
        return TYPE_LEFT;
    }

    return TYPE_NONE;
}

得分取决于正确程度和总共花的时间。

链接时如何选择C Run-Time(CRT) library

Michael — Thu, 30 Jun 2005 04:48:00 GMT

在Windows下有六种类型CRTLib(C运行库）:
Reusable Library                Switch    Library    Macro(s) Defined
----------------------------------------------------------------
Single Threaded                     /ML       LIBC               (none)
Static MultiThread                  /MT       LIBCMT         _MT
Dynamic Link (DLL)              /MD       MSVCRT        _MT and _DLL
Debug Single Threaded           /MLd      LIBCD          _DEBUG
Debug Static MultiThread        /MTd      LIBCMTD    _DEBUG and _MT
Debug Dynamic Link (DLL)    /MDd      MSVCRTD    _DEBUG, _MT, and _DLL

MT和MD都适用于多线程，其区别是：
MT为静态链接CRT，这样编译出来exe是自包含的，所以会相对大一些，但运行时不用再load CRT库。
MD为动态链接CRT，编译出来exe会小一些，运行时需要load CRT，性能有一点点损失。

任何工程都应该使用同样的CRT Library。即要么都是/ML，要么都是/MTD, 如此类推。

如果一个程序中混合使用不同类型的CRT，有时可以通过link，这样会存在不同CRT的copy，并导致以下问题：
    1）在一个lib中new出来内存，在另一个lib中delete，会crash。
    2）不能在多个lib中共享file handle。
    3）一个lib中设置locale（本地化有关），不能在另一个lib中起作用。

当工程比较大，包含的lib很多，特别当有外部lib（Third party library）存在时，link很容易发生下面这样的错误。
LIBCMTD.lib(chsize.obj) : error LNK2005: __chsize already defined in MSVCRTD.lib(MSVCRTD.dll)
这说明，你的工程使用了不同类型的CRT。这个时候首先一定要坚信这个原则：整个工程用同样的CRT Lib就可以解决问题。然后耐心一一检查每个lib。
如果恰恰某个外部lib用MT，另一个用MD，这个时候就比较痛苦。
如果有他们源码，就编译一个MT or MD类型的lib，以统一使用一个类型CRT。
如果没有，那可能只好选择其他的lib。

一般，我个人习惯将所有lib都编译为MT。出于其他需要，会同时提供MT(MTd)和MD(MDd)两种版本。

为什么Microsoft code sample倾向使用ZeroMemory而不是{0}?

Michael — Wed, 29 Jun 2005 13:23:00 GMT

这是最近看的一片短文的title，当时就很好奇。
经常查阅MSDN的程序员都会有这个印象，微软code sample中常见的是ZeroMemory，而不是语言提供的“{0}”清零功能（不过，我一直也没有问个why）。c++语法中声明对数组或纯结构（struct），可以使用例如SPerson sTest = {0}; 来将所有成员置0.

那篇文章的解释是，Microsoft使用ZeroMemory会更clear，因为“= {0}” 这样的语法有些生僻，不是所有人都可以一下子明白。

实际上，两者还是有一些区别。
其一，ZeroMemory会将结构所有字节置0，而={0}只会将成员置0，其中padding字节不变。

其二，但一个struct有构造函数或虚函数时，ZeroMemory可以，而={0}会编译不过。显然，后者起到了一些保护作用，因为对一个有虚函数的对象使用ZeroMemory时，会将其虚函数的指针置0，这是非常危险的，因为调用虚函数时，程序显然会crash。参看如下代码：

struct SPerson
{
    //SPerson(){    }
    char c;
    float s;
};

class CTestVirtual
{
public:
    CTestVirtual()
    {
    }

    virtual int Draw()
    {
        return 10;
    }

    int a;
};

void Test()
{
    char sztmp[20];
    ZeroMemory(sztmp, sizeof(sztmp));

    SPerson sTest = {0};
    int i = sizeof(SPerson);

    //CTestVirtual otv = {0};    //Compire error
    CTestVirtual tv;
    ZeroMemory(&tv, sizeof(tv));
    tv.Draw();        //As it is an object, don't use the virtual function pointer, so don't crash.
    CTestVirtual *pTv = &tv;
    pTv->Draw();    //Crash!!!
}

因此，在windows平台下，对于数组或纯结构使用ZeroMemory是安全的，对于class，则使用构造函数，不要调用ZeroMemory。如果有跨平台要求，使用＝{0}则可以减少一些工作。

标准输出(stdout)重定向

Michael — Mon, 06 Jun 2005 10:28:00 GMT

公司有个同事用了一个很老的DLL，其信息都是输出到控制台窗口（console），问能不能将其输出到MFC的view中？

我们知道，标准输出默认情况下是输出到console中。
要将stdout重定向到文件, 比较简单，code如下（MSDN中源码）：

//将stdout重定向到文件中
   FILE *stream ;
   if((stream = freopen("file.txt", "w", stdout)) == NULL)
      exit(-1);
   printf("this is stdout output ");

   //重新将stdout重定向到console
   stream = freopen("CON", "w", stdout);
   printf("And now back to the console once again ");

但是将stdout重定向某个MFC程序View中显示出来，实现稍有点复杂。总体思路是
1）创建一个匿名管道(pipe)
2）将stdout重定向到这个pipe上
3）创建一个线程，读取pipe中的内容，然后将这些内容输出到你希望输出的地方，譬如一个view里。
启动进程时，先调用StdoutRedirect()，那么这个进程中所有的printf都可以输出到View里。

code如下：

HANDLE hOutputReadTmp = 0;
HANDLE hOutputWrite = 0;
void DisplayError(char* szMsg)
{
    TRACE(szMsg);
    TRACE(" ");
}

DWORD WINAPI ReadStdout(LPVOID lpvThreadParam)
{
  CHAR lpBuffer[256];
  DWORD nBytesRead;

  while(TRUE)
  {
     if (!ReadFile(hOutputReadTmp, lpBuffer,sizeof(lpBuffer),
                                      &nBytesRead,NULL) || !nBytesRead)
     {
        if (GetLastError() == ERROR_BROKEN_PIPE)
           break; // pipe done - normal exit path.
        else
           DisplayError("ReadFile"); // Something bad happened.
     }

     if (nBytesRead >0 )
     {
         //获取到printf的输出
         lpBuffer[nBytesRead] = 0;

        //replace your code here. Do it yourself.
        DisplayInView(lpBuffer);
     }
  }

  return 0;
}

void StdoutRedirect()
{
    if (!CreatePipe(&hOutputReadTmp,&hOutputWrite,0,0))
         DisplayError("CreatePipe");

   int hCrt;
   FILE *hf;
   //AllocConsole();
   hCrt = _open_osfhandle(
             (long)hOutputWrite,
             _O_TEXT
          );
   hf = _fdopen( hCrt, "w" );
   *stdout = *hf;
   int i = setvbuf( stdout, NULL, _IONBF, 0 );

      // Launch the thread that gets the input and sends it to the child.
    DWORD ThreadID;
    HANDLE hThread = ::CreateThread(NULL,0,ReadStdout,
                              0,0,&ThreadID);
}

最后，顺便提一下，如果要redirect某个console进程的stdout，也是类似。MSDN有详细例子。搜索“stdout redirect”即可。