Question

我正在研究一个我正在为一个班级设计的项目。我想尝试模仿菲利普斯的“深渊”或常见的自制“BobLight”的行为。它通过屏幕截图得到屏幕的平均颜色，在我的例子中，逐像素地平均RGB像素数据。我决定尝试通过使用Windows GDI api来做到这一点，到目前为止我得到了一些相当不错的结果，我学会了不使用GetPixel（）来读取像素数据，因为它非常慢，而且我的速度完全可以接受所以远。

我的问题是，当我尝试在一个可以运行所需的循环中实现它时，它最终会挂起，给出一个错误说：std :: bad_alloc在内存位置0x0027F9C4。

在这行代码中：

BYTE *lpbitmap = new BYTE[dwBmpSize];

我在下面附上了我的整个代码有谁知道可能导致此错误的原因是什么？我不应该没有任何记忆，我真的不知所措。在此先感谢您的帮助！

#define _WIN32_WINNT    0x0501        //xp
#include <windows.h>
#include <iostream>
#include <vector>
#include <time.h>

#define NUM_FRAMES 180

using namespace std;


int main()
{ 
    int time_start, time_finish;
    float time_total;
    time_start = clock();

for(int z = 0; z < NUM_FRAMES; z++)
{

HDC hScreenDC = CreateDC(TEXT("DISPLAY"), NULL, NULL, NULL);     
// and a device context to put it in
HDC hMemoryDC = CreateCompatibleDC(hScreenDC);

int x = GetDeviceCaps(hScreenDC, HORZRES);
int y = GetDeviceCaps(hScreenDC, VERTRES);

HBITMAP hBitmap = CreateCompatibleBitmap(hScreenDC, x, y);

// get a new bitmap
HBITMAP hOldBitmap = (HBITMAP)SelectObject(hMemoryDC, hBitmap);


RECT desktop;
// Get a handle to the desktop window
const HWND hDesktop = GetDesktopWindow();
// Get the size of screen to the variable desktop
GetWindowRect(hDesktop, &desktop);
// The top left corner will have coordinates (0,0)
// and the bottom right corner will have coordinates
// (horizontal, vertical)
int horizontal = desktop.right;
int vertical = desktop.bottom;


BitBlt(hMemoryDC, 0, 0, horizontal, vertical, hScreenDC, 0, 0, SRCCOPY);
hBitmap = (HBITMAP)SelectObject(hMemoryDC, hOldBitmap);

BITMAPINFOHEADER   bi;

bi.biSize = sizeof(BITMAPINFOHEADER);    
bi.biWidth = horizontal;    
bi.biHeight = vertical;  
bi.biPlanes = 1;    
bi.biBitCount = 32;    
bi.biCompression = BI_RGB;    
bi.biSizeImage = 0;  
bi.biXPelsPerMeter = 0;    
bi.biYPelsPerMeter = 0;    
bi.biClrUsed = 0;    
bi.biClrImportant = 0;

DWORD dwBmpSize = ((horizontal * bi.biBitCount + 31) / 32) * 4 * vertical;

// Starting with 32-bit Windows, GlobalAlloc and LocalAlloc are implemented as wrapper     functions that 
// call HeapAlloc using a handle to the process's default heap. Therefore, GlobalAlloc     and LocalAlloc 
// have greater overhead than HeapAlloc.
HANDLE hDIB = GlobalAlloc(GHND,dwBmpSize); 
BYTE *lpbitmap = new BYTE[dwBmpSize]; 



//BITMAP bm;
GetDIBits(hMemoryDC,hBitmap,0,(UINT)vertical,lpbitmap, (BITMAPINFO *)&bi, DIB_RGB_COLORS);
GetObject (hBitmap, sizeof(lpbitmap), &lpbitmap);


unsigned long red = 0, green = 0, blue = 0;

for (int i = 0; i < horizontal; i++)
{
    for (int j = 0; j < vertical; j++)
    {
    blue += lpbitmap[0];
    green += lpbitmap[1];
    red += lpbitmap[2];

    lpbitmap += 4;
    }
}

red = red/(horizontal*vertical);
green = green/(horizontal*vertical);
blue = blue/(horizontal*vertical);

//cout << "Red: " << red << "    Green: " << green << "    Blue: " << blue << endl;
}
time_finish = clock();
time_total = time_finish - time_start;

cout << endl << NUM_FRAMES/((time_total)/10000) << endl;
    //release
    //DeleteDC(hdc);
    //DeleteObject(hbmp);
    //ReleaseDC(NULL, hdcScreen);

        std::system("pause");
    return 0;
}

Answer 1

您的发布代码似乎已被注释掉，这意味着您每帧至少会泄漏hBitmap，这会快速累加（特定时间约为470MB / s）。实际上，看起来你也在泄漏lpbitmap，因此你的180帧将在1080p显示器上刻录大约2.8 GB。

使用c ++和GDI分析屏幕截图以获得屏幕的平均颜色

1 个答案: