使用WinApi和C++仅截取两个显示器的多显示器截图

Question

我有一个使用WinApi和C++在Windows平台上进行截屏的函数。它在一个或两个监视器上运行得很好，但当我在具有3个或更多监视器的机器上运行它时，它只拍摄了两个监视器的图片。

我认为我的问题是“主”监视器左侧的监视器内容被切断了。不幸的是，我不知道如何修复它或我做错了什么。

我已经阅读了BitBlt和StretchBlt的相关资料，并尝试使用它们，但没有成功。

这就是我正在做的事情:

// Get the system metrics
const int width = GetSystemMetrics(SM_CXVIRTUALSCREEN);
const int height = GetSystemMetrics(SM_CYVIRTUALSCREEN);

对于所有的显示器，宽度和高度都能完美计算。

// Create a normal DC and a memory DC for the entire screen. The normal DC provides a "snapshot" of the screen contents.
// The memory DC keeps a copy of this "snapshot" in the associated bitmap.
const HDC hdcScr = CreateDCW(TEXT("DISPLAY"), NULL, NULL, NULL);
const HDC hdcMem = CreateCompatibleDC(hdcScr);



if (!StretchBlt(hdcMem, 0, 0, width, height, hdcScr, 0, 0, width, height, SRCCOPY)) {
    return;
}


// Create a compatible bitmap for hdcScreen.
const HBITMAP hbmScr = CreateCompatibleBitmap(hdcScr, width, height);
if (hbmScr == 0) {
    return;
}

if (!BitBlt(hdcMem, 0, 0, width, height, hdcScr, 0, 0, SRCCOPY)) {
    return;
}



// Select the bitmaps into the compatible DC.
if (!SelectObject(hdcMem, hbmScr)) {
    DeleteDC(hdcScr);
    DeleteDC(hdcMem);
    DeleteObject(hbmScr);
    return;
}

// Copy color data for the entire display into a bitmap that is selected into a compatible DC.
if (!StretchBlt(hdcMem, 0, 0, width, height, hdcScr, 0, 0, width, height, SRCCOPY)) {
    DeleteDC(hdcScr);
    DeleteDC(hdcMem);
    DeleteObject(hbmScr);
    return;
}

BITMAP bmp;

// Retrieve the bitmap's color format, width, and height.
if (!GetObject(hbmScr, sizeof(BITMAP), reinterpret_cast<LPSTR>(&bmp))) {
    DeleteDC(hdcScr);
    DeleteDC(hdcMem);
    DeleteObject(hbmScr);
    return;
}


// Convert the color format to a count of bits.
unsigned short cClrBits = bmp.bmPlanes * bmp.bmBitsPixel;
if (cClrBits == 1) {
    cClrBits = 1;
}
else if (cClrBits <= 4) {
    cClrBits = 4;
}
else if (cClrBits <= 8) {
    cClrBits = 8;
}
else if (cClrBits <= 16) {
    cClrBits = 16;
}
else if (cClrBits <= 24) {
    cClrBits = 24;
}
else {
    cClrBits = 32;
}


PBITMAPINFO pbmi;

// Allocate memory for the BITMAPINFO structure. (This structure contains a BITMAPINFOHEADER structure and an array of RGBQUAD data structures.)
if (cClrBits != 24) {
    pbmi = static_cast<PBITMAPINFO>(LocalAlloc(LPTR, sizeof(BITMAPINFOHEADER) + sizeof(RGBQUAD) * (1 << cClrBits)));
}
else { // There is no RGBQUAD array for the 24-bit-per-pixel format.
    pbmi = static_cast<PBITMAPINFO>(LocalAlloc(LPTR, sizeof(BITMAPINFOHEADER)));
}

// Initialize the fields in the BITMAPINFO structure.
pbmi->bmiHeader.biSize = sizeof(BITMAPINFOHEADER);
pbmi->bmiHeader.biWidth = bmp.bmWidth;
pbmi->bmiHeader.biHeight = bmp.bmHeight;
pbmi->bmiHeader.biPlanes = bmp.bmPlanes;
pbmi->bmiHeader.biBitCount = bmp.bmBitsPixel;

if (cClrBits < 24) {
    pbmi->bmiHeader.biClrUsed = (1 << cClrBits);
}

// If the bitmap is not compressed, set the BI_RGB flag.
pbmi->bmiHeader.biCompression = BI_RGB;

// Compute the number of bytes in the array of color indices and store the result in biSizeImage.
pbmi->bmiHeader.biSizeImage = (pbmi->bmiHeader.biWidth + 7) / 8 * pbmi->bmiHeader.biHeight * cClrBits;

// Set biClrImportant to 0, indicating that all of the device colors are important.
pbmi->bmiHeader.biClrImportant = 0;


const PBITMAPINFOHEADER pbih = reinterpret_cast<PBITMAPINFOHEADER>(pbmi);              // bitmap info-header
const LPBYTE lpBits = static_cast<LPBYTE>(GlobalAlloc(GMEM_FIXED, pbih->biSizeImage)); // memory pointer

if (!lpBits) {
    DeleteDC(hdcScr);
    DeleteDC(hdcMem);
    LocalFree(pbmi);
    GlobalFree(lpBits);
    DeleteObject(hbmScr);
    return;
}

// Retrieve the color table (RGBQUAD array) and the bits (array of palette indices) from the DIB.
if (!GetDIBits(hdcMem, hbmScr, 0, pbih->biHeight, lpBits, pbmi, DIB_RGB_COLORS)) {
    DeleteDC(hdcScr);
    DeleteDC(hdcMem);
    LocalFree(pbmi);
    GlobalFree(lpBits);
    DeleteObject(hbmScr);
    return;
}

BITMAPFILEHEADER hdr; // bitmap file-header

hdr.bfType = 0x4d42; // ('M' << 8) + 'B';

// Calculate the size of the entire file.
hdr.bfSize = sizeof(BITMAPFILEHEADER) + pbih->biSize + pbih->biClrUsed * sizeof(RGBQUAD) + pbih->biSizeImage;
hdr.bfReserved1 = NULL;
hdr.bfReserved2 = NULL;

// Calculate the offset to the array of color indices.
hdr.bfOffBits = sizeof(BITMAPFILEHEADER) + pbih->biSize + pbih->biClrUsed * sizeof(RGBQUAD);


const DWORD cb = pbih->biSizeImage; // incremental count of bytes


std::stringstream outputBitmap;

// Write the BITMAPFILEHEADER into the .BMP file.
outputBitmap.write( reinterpret_cast<LPSTR>(&hdr), sizeof(BITMAPFILEHEADER));

// Write the BITMAPINFOHEADER and RGBQUAD array into the file.
outputBitmap.write(reinterpret_cast<LPSTR>(pbih), sizeof(BITMAPINFOHEADER) + pbih->biClrUsed * sizeof(RGBQUAD));

// Write the array of color indices
outputBitmap.write(reinterpret_cast<LPSTR>(lpBits), cb);




// To test the whole thing
std::ofstream out("test.bmp", std::ios::out | std::ios::binary);
out << outputBitmap.str();
out.close();


// Cleanup
DeleteDC(hdcScr);
DeleteDC(hdcMem);
LocalFree(pbmi);
GlobalFree(lpBits);
DeleteObject(hbmScr);

- The Coder

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如果显示器位于原点左侧，那么显示DC是否不需要负坐标（在您的情况下为负x）？ - Chris Becke

在 Stack Overflow 上可能有数百万的屏幕截图问题，但仍然不断出现。 - Jonathan Potter

Answer 1

显示器的坐标可能是负数，所以假设（0,0）是显示器左上角是危险的^†。真正的原点（x，y）由系统度量SM_XVIRTUALSCREEN和SM_YVIRTUALSCREEN给出。然后，您需要更新所有BLT以引用正确的源位置。这使得函数的第一部分：

int x = GetSystemMetrics(SM_XVIRTUALSCREEN);
int y = GetSystemMetrics(SM_YVIRTUALSCREEN);
int w = GetSystemMetrics(SM_CXVIRTUALSCREEN);
int h = GetSystemMetrics(SM_CYVIRTUALSCREEN);

BOOL ok = StretchBlt(hdcMem, 0, 0, w, h, hdcScr, x, y, w, h, SRCCOPY);
// And So On...

另外，由于这类问题经常被问到，您可以使用类似以下不完全生产质量的代码来获取单个监视器：

#define WIN32_LEAN_AND_MEAN
#include <Windows.h>
#include <iostream>
using std::wcout;
using std::endl;

typedef struct tagMonData
{
    int current;
    MONITORINFOEXW* info;
} MonData;

BOOL EnumProc(HMONITOR hMonitor, HDC hdcMonitor, LPRECT lprcMonitor, LPARAM dwData)
{
    MonData* data = (MonData*)dwData;
    data->info[data->current].cbSize = sizeof(MONITORINFOEXW);
    return GetMonitorInfoW(hMonitor, &(data->info[data->current++]));
}

BOOL GetAllMonitorInfo(MonData* data)
{
    return EnumDisplayMonitors(NULL, NULL, (MONITORENUMPROC)(&EnumProc), (LPARAM)(data));
}

int main()
{
    int cMonitors = GetSystemMetrics(SM_CMONITORS);
    MonData data;
    data.current = 0;
    data.info = (MONITORINFOEXW*)calloc(cMonitors, sizeof(MONITORINFOEXW));

    if (!GetAllMonitorInfo(&data)) return 1;

    for (int i = 0; i < cMonitors; i++)
    {
        wcout << data.info[i].szDevice << "X: " << data.info[i].rcMonitor.left << " Y: " << data.info[i].rcMonitor.top << endl;
    }

    free(data.info);
    return 0;
}

请注意，如果 GetMonitorInfoW 调用失败，则会捷径并留下不完整的信息。

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^† Raymond Chen 在 The Old New Thing 上写了几篇关于 Windows 坐标系统复杂性的文章。

- theB