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請幫助這個初學者的程序獲取gdk屏幕屬性。我有一個小型C++程序來查找連接的顯示單元。我在Linux Debian上使用C++。 gdk_screen_get_default()
不返回Screen
對象。如果我不檢查屏幕對象,則會發生以下錯誤。gdk_screen_get_default()返回null
錯誤 (進程:8023):GDK-CRITICAL **:gdk_screen_get_monitor_geometry:斷言 'GDK_IS_SCREEN(屏幕)' 失敗
我通過相關帖子去了,簡稱爲this以下代碼片段。
感謝您的幫助。任何解決此問題的指針/指南都會有所幫助。
我有一臺顯示器連接和顯示設置
$ echo $XDG_CURRENT_DESKTOP
GNOME
$ echo $DISPLAY
:0
CODE
#include <gdk/gdk.h>
#include <iostream>
/*
GTK version 3.14.5
g++ getScreenInfo.cpp -o getScreenInfo `pkg-config gtk+-3.0 --cflags --libs`
*/
int main()
{
GdkScreen *screen;
screen = gdk_screen_get_default();
int num_monitors;
int i;
if (screen)
{
num_monitors = gdk_screen_get_n_monitors(screen);
for (i = 0; i < num_monitors; i++)
{
GdkRectangle rect;
gdk_screen_get_monitor_geometry (screen, i, &rect);
std::cout << "monitor " << i << ": coordinates (" << rect.x << ","
<< rect.y << ", size (" << rect.width << "," << rect.height << ")"
<< std::endl;
}
}else
{
std::cout << "Couldn't obtain default screen object" << std::endl;
}
}
2017年4月27日編輯:議決
#include <iostream>
#include <gdk/gdk.h>
#include <gtk/gtk.h>
/*
GTK version 3.14.5
To compile:
g++ getScreenInfo.cpp -o getScreenInfo `pkg-config gtk+-3.0 --cflags --libs`
*/
int main(int argc, char *argv[])
{
gtk_init(&argc, &argv);
GdkScreen *screen = gdk_screen_get_default();
int num_monitors;
int i;
if (screen)
{
num_monitors = gdk_screen_get_n_monitors(screen);
for (i = 0; i < num_monitors; i++)
{
GdkRectangle rect;
gdk_screen_get_monitor_geometry (screen, i, &rect);
std::cout << "monitor " << i << ": offsets (" << rect.x << ","
<< rect.y << ", size (" << rect.width << "," << rect.height << ")"
<< std::endl;
}
}
else
{
std::cout << "Couldn't obtain default screen object" << std::endl;
}
// To query primary display properties
guint monitor = gdk_screen_get_primary_monitor(screen);
GdkRectangle screen_geometry = { 0, 0, 0, 0 };
gdk_screen_get_monitor_geometry(screen, monitor, &screen_geometry);
std::cout << screen_geometry.x << std::endl;
std::cout << screen_geometry.y << std::endl;
std::cout << screen_geometry.width << std::endl;
std::cout << screen_geometry.height << std::endl;
}
請將問題標記爲接受而不是編輯標題。謝謝! :) – Alfabravo