我写的小程序

RGB24转YV12(按公式转，没有任何优化)

玻璃小屋 — Thu, 12 Nov 2009 07:52:00 GMT

// 最直接的(没有任何优化，直接按公式转换！)把RGB24转成YV12的算法
// 参数：宽(w) 和高(h) 必须是 4对齐的！不然会有错误！
// 输出结果 YV12 格式：
// y01 y02 y03 y04 -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// v01 v02 -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -
// u01 u02 -   -   -   -   -   -   -   -   -   -
// -   -   -   -   -   -   -   -   -   -   -   -

typedef struct tagRGB24{
        unsigned char    b;
        unsigned char    g;
        unsigned char    r;
} RGB24;

void HotRGB24_2_YV12(unsigned char* yv12,unsigned char* rgb24,int w,int h)
{
    int iBufLen = w * h;
    int i,j,vay,vau,vav;
    unsigned char* cv;        // 当前坐标的v(current v);
    unsigned char* nv;        // 在cv下一行的对应位置的 v;
    unsigned char* cu;        // 当前坐标的u(current u);
    unsigned char* nu;        // 在cu下一行的对应位置的 u;
    unsigned char v01,v02,v11,v12,u01,u02,u11,u12;            // 需要整合的相邻的4个象素如下
    /*
    v01 v02 - - - -
    v11 v12 - - - -
    -   -   - - - -
    u01 u02
    u11 u12
    */
    unsigned char* vv = new unsigned char[iBufLen];            // 每个RGB单位对应的V!
    unsigned char* uu = new unsigned char[iBufLen];            // 每个RGB单位对应的U!

    // 按标准算法从RGB24计算出所有YUV

    RGB24 * pRGB = (RGB24*)rgb24;
    unsigned char* y = yv12;                                // 这里直接用 yuv 缓冲。。省了copy了。
    unsigned char* v = vv;
    unsigned char* u = uu;

    for(i = 0; i < h; i++)
    {
        for(j = 0; j < w; j++)
        {

            //Y =   0.299R + 0.587G + 0.114B
            //U =  -0.147R - 0.289G + 0.436B
            //V =   0.615R - 0.515G - 0.100B

            // 这个算法颜色不正啊。。
            //vay =  pRGB->r *  0.299 + pRGB->g *  0.587 + pRGB->b *  0.114;    // 根据公式计算出Y
            //vav =  pRGB->r *  0.615 + pRGB->g * -0.515 + pRGB->b * -0.100;    // 根据公式计算出V
            //vau =  pRGB->r * -0.147 + pRGB->g * -0.289 + pRGB->b *  0.436;    // 根据公式计算出U

            //Y = round( 0.256788 * R + 0.504129 * G + 0.097906 * B) +  16
            //U = round(-0.148223 * R - 0.290993 * G + 0.439216 * B) + 128
            //V = round( 0.439216 * R - 0.367788 * G - 0.071427 * B) + 128

            // 好象这个算法颜色正，而且是MSDN中列出的算法
            vay =  0.256788 * pRGB->r + 0.504129 * pRGB->g + 0.097906 * pRGB->b +  16;
            vau = -0.148223 * pRGB->r - 0.290993 * pRGB->g + 0.439216 * pRGB->b + 128;
            vav =  0.439216 * pRGB->r - 0.367788 * pRGB->g - 0.071427 * pRGB->b + 128;

            *y =  vay < 0 ? 0 : (vay > 255 ? 255: vay);                        // 如果Y小于0置换成0，如果Y大于255就置换成255
            *v =  vav < 0 ? 0 : (vav > 255 ? 255: vav);                        // 如果V小于0置换成0，如果V大于255就置换成255
            *u =  vau < 0 ? 0 : (vau > 255 ? 255: vau);                        // 如果U小于0置换成0，如果U大于255就置换成255

            y++;    // 移动到下一位！
            v++;
            u++;
            pRGB++;
        }
    }

    v = yv12 + iBufLen;                                // 记录成品YV12的V的位置
    u = v    + (iBufLen >> 2);                        // 记录成品YV12的U的位置

    for(i = 0; i < h; i+=2)                            // 由于 V 和 U 只记录隔行的，所以 += 2;
    {
        cv = vv + i * w;                            // 取得第i     行的v;
        nv = vv + (i + 1) * w;                        // 取得第i + 1 行的v

        cu = uu + i * w;                            // 取得第i     行的u;
        nu = uu + (i + 1) * w;                        // 取得第i + 1 行的u

        for(j = 0; j < w; j+=2)                        // 由于一躺循环我们访问两个 uu 或 vv 所以 += 2;
        {
            v01 = *(cv + j);                        // 取得第i     行的第j     个v的具体的值
            v02 = *(cv + j + 1);                    // 取得第i     行的第j + 1 个v的具体的值
            v11 = *(nv + j);                        // 取得第i + 1 行的第j     个v的具体的值
            v12 = *(nv + j + 1);                    // 取得第i + 1 行的第j + 1 个v的具体的值

            *v = (v01 + v02 + v11 + v12) / 4;        // 取v01,v02,v11,v12的平均值给v

            u01 = *(cu + j);                        // 取得第i     行的第j     个u的具体的值
            u02 = *(cu + j + 1);                    // 取得第i     行的第j + 1 个u的具体的值
            u11 = *(nu + j);                        // 取得第i + 1 行的第j     个u的具体的值
            u12 = *(nu + j + 1);                    // 取得第i + 1 行的第j + 1 个u的具体的值

            *u = (u01 + u02 + u11 + u12) / 4;        // 取u01,u02,u11,u12的平均值给u

            v++;    // 移动到下一位！
            u++;

        }
    }

    delete [] vv;
    delete [] uu;

}

WSARecv 投递缓冲区也要对齐？？？不然就10014!

玻璃小屋 — Tue, 06 May 2008 17:23:00 GMT

如果把下面代码中背景色为绿色部分的 int iAlign = 4; 其中的 4 改为1或2，WSARecv就会返回 10014 错误。看来WSARecv 的投递的缓冲区也是要求对齐的。。正是因为这个问题。。浪费了我一天的时间调试我的内存池代码。。5555555555

#include <stdio.h>
#include "winsock2.h"

#define DATA_BUFSIZE 4096

void main() {
  //-----------------------------------------
  // Declare and initialize variables
  WSABUF DataBuf;
  char buffer[DATA_BUFSIZE];
  DWORD EventTotal = 0,
    RecvBytes = 0,
    Flags = 0,
    BytesTransferred = 0,
    CallBack = 0;
  WSAEVENT EventArray[WSA_MAXIMUM_WAIT_EVENTS];
  WSAOVERLAPPED AcceptOverlapped;
  SOCKET ListenSocket, AcceptSocket;

  //-----------------------------------------
  // Initialize Winsock
  WSADATA wsaData;
  WSAStartup(MAKEWORD(2,2), &wsaData);

  //-----------------------------------------
  // Create a listening socket bound to a local
  // IP address and the port specified
  ListenSocket = socket(AF_INET, SOCK_STREAM, IPPROTO_TCP);
  u_short port = 27015;
  char* ip;
  sockaddr_in service;
  service.sin_family = AF_INET;
  service.sin_port = htons(port);
  hostent* thisHost;
  thisHost = gethostbyname("");
  ip = inet_ntoa (*(struct in_addr *)*thisHost->h_addr_list);

  service.sin_addr.s_addr = inet_addr(ip);

  //-----------------------------------------
  // Bind the listening socket to the local IP address
  // and port number
  bind(ListenSocket, (SOCKADDR *) &service, sizeof(SOCKADDR));

  //-----------------------------------------
  // Set the socket to listen for incoming
  // connection requests
  listen(ListenSocket, 1);
  printf("Listening ");

  //-----------------------------------------
  // Accept and incoming connection request
  AcceptSocket = accept(ListenSocket, NULL, NULL);
  printf("Client Accepted ");

  //-----------------------------------------
  // Create an event handle and setup an overlapped structure.
  EventArray[EventTotal] = WSACreateEvent();
  ZeroMemory(&AcceptOverlapped, sizeof(WSAOVERLAPPED));
  AcceptOverlapped.hEvent = EventArray[EventTotal];
  DataBuf.len = DATA_BUFSIZE;
  DataBuf.buf = buffer;
  EventTotal++;

  LPWSABUF lpwsaBuf;

  //int iAlign = 1;
  int iAlign = 4;

  lpwsaBuf = (LPWSABUF) ((new CHAR[sizeof(WSABUF) + iAlign]) + iAlign);//pOverlapBuff->GetWSABuffer();
  lpwsaBuf->buf = new CHAR[52];
  lpwsaBuf->len = 52;

  //-----------------------------------------
  // Call WSARecv to receive data into DataBuf on
  // the accepted socket in overlapped I/O mode
  if (WSARecv(AcceptSocket, lpwsaBuf, 1, &RecvBytes, &Flags, &AcceptOverlapped, NULL) == SOCKET_ERROR)
  {
      if (WSAGetLastError() != WSA_IO_PENDING)
      {
          printf("Error occured at WSARecv() %d ",WSAGetLastError());
          closesocket(AcceptSocket);
          return;
      }
  }

  //-----------------------------------------
  // Process overlapped receives on the socket
  while (1) {
    DWORD Index;

    //-----------------------------------------
    // Wait for the overlapped I/O call to complete
    Index = WSAWaitForMultipleEvents(EventTotal, EventArray, FALSE, WSA_INFINITE, FALSE);

    //-----------------------------------------
    // Reset the signaled event
    WSAResetEvent(EventArray[Index - WSA_WAIT_EVENT_0]);

    //-----------------------------------------
    // Determine the status of the overlapped event
    WSAGetOverlappedResult(AcceptSocket, &AcceptOverlapped, &BytesTransferred, FALSE, &Flags);

    //-----------------------------------------
    // If the connection has been closed, close the accepted socket
    if (BytesTransferred == 0) {
      printf("Closing Socket %d ", AcceptSocket);
      closesocket(AcceptSocket);
      WSACloseEvent(EventArray[Index - WSA_WAIT_EVENT_0]);
      return;
    }

    //-----------------------------------------
    // If data has been received, echo the received data
    // from DataBuf back to the client
    if (WSASend(AcceptSocket, &DataBuf, 1, &RecvBytes, Flags, &AcceptOverlapped, NULL) == SOCKET_ERROR)
      printf("WSASend() is busted ");

    //-----------------------------------------
    // Reset the changed flags and overlapped structure
    Flags = 0;
    ZeroMemory(&AcceptOverlapped, sizeof(WSAOVERLAPPED));

    AcceptOverlapped.hEvent = EventArray[Index - WSA_WAIT_EVENT_0];

    //-----------------------------------------
    // Reset the data buffer
    DataBuf.len = DATA_BUFSIZE;
    DataBuf.buf = buffer;
  }
}

结构对齐对结构成员的访问效率影响的测试。

玻璃小屋 — Fri, 02 May 2008 14:14:00 GMT

测试一下结构对齐对其成员访问效率的影响，程序中应该涉及到了对结构成员变量的读/写操作，测试应该是比较全面可信的。
测试结果表明，改变结构默认的对齐方式对其成员访问效率的影响是蛮大的。另外按CPU宽度来测试似乎与默认对齐方式的结果差不多，看来 #pragma pack(1)还是要慎用！！唉~~ 不过要是在 pack(1) 慎用的情况下，做socket 程序时。。。发送或接收数据的结构可怎么写？？用 pack(1) 定义socket的收/发数据结构用起来很方便啊。。可是如果为了效率而舍弃的啊？这些数据该规划用什么方式来访问呢？有比用 pack(1)来定义结构更方遍，更高效的方法吗？

#pragma pack(1)

class Test
{
public:
    char    a;
    int        b;
};

#pragma pack()

class Test1
{
public :
    char a;
    int b;
};

__declspec(align(32)) class S
{
public:
     char a;
     int b;
} ;

void main()
{

    for(int k = 0; k < 20; k++)
    {
        Test t;
        t.a = 0x33;
        t.b = 3;
        int j = 0;
        int g = 0;
        DWORD dTick = GetTickCount();
        for(int i = 0; i < 100000000; i++)
        {
            if(t.b == g)
            {
                j++;
            }

            t.b = i;
            g = j;
        }

        printf("Test0：%d %d %d ",__alignof(Test),sizeof(Test),GetTickCount() - dTick);

        Test1 t1;
        t1.a = 0x33;
        t1.b = 3;


        dTick = GetTickCount();
        for(int i = 0; i < 100000000; i++)
        {
            if(t1.b == g)
            {
                j++;
            }

            t1.b = i;
            g = j;
        }

        printf("Test1：%d %d %d ",__alignof(Test1),sizeof(Test1),GetTickCount() - dTick);

        S t2;
        t2.a = 0x33;
        t2.b = 3;

        dTick = GetTickCount();
        for(int i = 0; i < 100000000; i++)
        {
            if(t2.b == g)
            {
                j++;
            }

            t2.b = i;
            g = j;
        }

        printf("Test2：%d %d %d ",__alignof(S),sizeof(S),GetTickCount() - dTick);

    }
}

输出结果：

Test0：1        5       437
Test1：4        8       125
Test2：32       32      125