c++ - 使用 CreateFile 打开显示时的正确路径是什么？

Question

最简单的例子:

#include <Windows.h>
#include <ntddvdeo.h>
#include <string>


std::string GetLastErrorAsString()
{
    LPSTR msgBuffer = nullptr;
    DWORD errorMessageID = ::GetLastError();
    std::string msg("No Error");

    if(errorMessageID != 0)
    {
        size_t size = FormatMessageA(FORMAT_MESSAGE_ALLOCATE_BUFFER |
                                     FORMAT_MESSAGE_FROM_SYSTEM |
                                     FORMAT_MESSAGE_IGNORE_INSERTS,
                                     NULL,
                                     errorMessageID,
                                     MAKELANGID(LANG_NEUTRAL, 
                                     SUBLANG_DEFAULT),
                                     (LPSTR)&msgBuffer,
                                     0,
                                     NULL);
        msg = std::string(msgBuffer, size);

        //Free buffer
        LocalFree(msgBuffer);
    }
    return msg;
}

int APIENTRY wWinMain(  _In_ HINSTANCE hInstance,
                        _In_opt_ HINSTANCE hPrevInstance,
                        _In_ LPWSTR    lpCmdLine,
                        _In_ int       nCmdShow)
{
    DWORD nOutBufferSize = 0;
    DWORD result = NULL;

    typedef struct _DISPLAY_BRIGHTNESS {
       UCHAR ucDisplayPolicy;
       UCHAR ucACBrightness;
       UCHAR ucDCBrightness;
    } DISPLAY_BRIGHTNESS, *PDISPLAY_BRIGHTNESS;

    DISPLAY_BRIGHTNESS displayBrightness;

    displayBrightness.ucDisplayPolicy = 0;
    displayBrightness.ucACBrightness = 50;
    displayBrightness.ucDCBrightness = 0;
    nOutBufferSize = sizeof(displayBrightness);

    HANDLE h = CreateFile(  L"\\\\.\\DISPLAY2",
                            GENERIC_READ | GENERIC_WRITE,
                            0, NULL,
                            OPEN_EXISTING,
                            0, NULL);

    if (h == INVALID_HANDLE_VALUE) {
        MessageBoxA(NULL, GetLastErrorAsString().c_str(), "Test", 
                    MB_OK | MB_ICONERROR);
        return false;
    }    
    return true;
}

我总是在 CreateFile 之后收到“访问被拒绝”，即使以管理员身份运行 Visual Studio 也是如此。所以问题是，如果 \\.\DISPLAY2 未被识别，Windows 还接受什么作为设备名称/路径？但是，这是我从 EnumDisplayMonitors 获得的确切设备名称:

Found Monitor with 1920x1080 called \\.\DISPLAY1 located at 0:00
Found Monitor with 1680x1050 called \\.\DISPLAY2 located at -1680:240

Answer 1

如果您想要物理监视器的句柄，请获取 HMONITOR来自 EnumDisplayMonitors 的回调并将其用作 GetPhysicalMonitorsFromHMONITOR 的参数.

这会给你 HANDLE在PHYSICAL_MONITOR目的。例如，您可以使用该句柄调用 these functions

但是，如前所述 - 也许您想要获得的是设备上下文的句柄？然后使用 CreateDC , CreateFile不会在这里工作。

---

编辑:

如果要更改亮度，请尝试第一种方法 - 获取物理显示器 HANDLE并使用 SetMonitorBrightness

编辑 2:

好的，这段代码对我有用，实际上是从 EnumDisplayMonitors 返回的设备路径对我也不起作用。我不是这段代码的作者，我找到了它 here (自己想不通)

DISPLAY_DEVICE ddAdapter;
ddAdapter.cb = sizeof(DISPLAY_DEVICE);
for (DWORD dwAdapterNum = 0; EnumDisplayDevices(NULL, dwAdapterNum, &ddAdapter, 0); dwAdapterNum++)
{
    if ((ddAdapter.StateFlags & (DISPLAY_DEVICE_ATTACHED_TO_DESKTOP | DISPLAY_DEVICE_MIRRORING_DRIVER)) == DISPLAY_DEVICE_ATTACHED_TO_DESKTOP)
    {
        DISPLAY_DEVICE ddMonitor;
        ddMonitor.cb = sizeof(DISPLAY_DEVICE);
        for (DWORD dwMonitorNum = 0; EnumDisplayDevices(ddAdapter.DeviceName, dwMonitorNum, &ddMonitor, EDD_GET_DEVICE_INTERFACE_NAME); dwMonitorNum++)
        {
            if (ddMonitor.StateFlags & DISPLAY_DEVICE_ACTIVE)
            {
                HANDLE hMonitor = CreateFile(ddMonitor.DeviceID, GENERIC_READ | GENERIC_WRITE, FILE_SHARE_READ, NULL, OPEN_EXISTING, 0, NULL);

                // ...

                CloseHandle(hMonitor);
            }
        }
    }
}