Question

我需要在多显示器系统中获取有关最近显示器的尺寸和坐标等信息到鼠标光标位置。我以前在Windows中完成过，我想知道如何在Linux X11中完成它。

使用下面的代码，我可以测量整个屏幕尺寸的总和，但不能单独测量每个显示器。

Screen *screen = DefaultScreenOfDisplay(DisplayHandle);
int xx = screen->width / 2 - Settings::WindowWidth / 2;
int yy = screen->height / 2 - Settings::WindowHeight / 2;

我之前的代码：

POINT mouse_position;
GetCursorPos(&mouse_position);
HMONITOR hMonitor = MonitorFromPoint(mouse_position, MONITOR_DEFAULTTOPRIMARY);
MONITORINFOEX monitor_info;
memset(&monitor_info, 0, sizeof(MONITORINFOEX));
monitor_info.cbSize = sizeof(MONITORINFOEX);
GetMonitorInfo(hMonitor, &monitor_info);

// CREATE WINDOW IN CENTER OF MONITOR //
int edge = GetSystemMetrics(SM_CXEDGE);
int fixed_frame = GetSystemMetrics(SM_CXFIXEDFRAME);
int monitor_width =  monitor_info.rcMonitor.right - monitor_info.rcMonitor.left;
int monitor_height =  monitor_info.rcMonitor.bottom - monitor_info.rcMonitor.top;
int xx = monitor_width / 2 - Settings::WindowWidth / 2;
int yy = monitor_height / 2 - Settings::WindowHeight / 2;
int win_x = xx - edge + monitor_info.rcMonitor.left;
int win_y = yy - fixed_frame + monitor_info.rcMonitor.top;

由于

Answer 1

如果您有两个构成单个桌面的显示器，请使用Xinerama扩展。下面的代码选择了可用显示器的最大屏幕，但你会明白这一点。

#include <X11/extensions/Xinerama.h>

// By default go fullscreen
m_winWidth = DisplayWidth (m_display, m_screenNo);
m_winHeight = DisplayHeight (m_display, m_screenNo);
// But, with Xinerama, use the largest physical screen
if (XineramaIsActive (m_display))
{
  int m = 0;
  int pixels = 0;

  XineramaScreenInfo *xs = XineramaQueryScreens (m_display, &m);

  if (0 != xs && m > 0)
  {
    for (int i = 0; i < m; i++)
    {
      //printf ("%dx%d, [%d, %d] %d\n", xs[i].width, xs[i].height, xs[i].x_org, xs[i].y_org, xs[i].screen_number);
      if (xs[i].width * xs[i].height > pixels)
      {
        m_xineramaScreen = xs[i].screen_number; // pick screen
        pixels = xs[i].width * xs[i].height;
        m_winWidth = xs[i].width;
        m_winHeight = xs[i].height;
      }
    }

    XFree (xs);
  }
}

在X11中获取有关最近的监视器到鼠标光标的信息

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