Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
QDesktopWidget * screen = QApplication::desktop();
QRect size = screen->availableGeometry(this);
The documentation says,
What is available will be subrect of screenGeometry() based on what the platform decides is available (for example excludes the Dock and Menubar on Mac OS X, or the taskbar on Windows).
What I want to know is how can I get the available screen geometry (without panels) on Linux? I tried above code on Linux, but it returns physical screen size.
For most of the cases
availableGeometry()works. I’m using virtual screen. Sometimes it doesn’t work properly.