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    Solved QRect and wrong rectangle coordinates

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    • J
      JulienD last edited by JulienD

      Hi,

      Currently using QT 5.15 and reading Qrect, I just want to make sure of one thing:

      QRect a(0,100,50,-50);
      

      yields (for topLeft, topRight, bottomRight, bottomLeft):

      QPoint(0,100)   QPoint(49,100)   QPoint(49,49)   QPoint(0,49)
      

      Which is correct (don't care about the -1)

      But if I do

      QRect a(0,100,50,50);
      

      it yields

      QPoint(0,100)   QPoint(49,100)   QPoint(49,149)   QPoint(0,149)
      

      Which is not logical since the topLeft is below the bottomLeft !
      So to be correct in all cases (rotation included), should I enforce the minus ? Or will another surprise arise?

      Thanks

      1 Reply Last reply Reply Quote 0
      • B
        Bonnie last edited by Bonnie

        What are you talking about?
        QPoint(0,149) is below QPoint(0,100) because the GUI coordinate system looks like this:
        https://doc.qt.io/qt-5/images/coordinatesystem-rect.png
        (picture is from: https://doc.qt.io/qt-5/coordsys.html)

        1 Reply Last reply Reply Quote 1
        • J
          JulienD last edited by

          Missed that info. Everything is alright. thanks!

          1 Reply Last reply Reply Quote 0
          • HoMa
            HoMa last edited by

            I want to add that the documentation of QRect reads different from your assumption: The consturctor which takes 4 int arguments assums "width" and "height" as last arguments. These are no point coordinates.

            QRect(int x, int y, int width, int height)
            
            1 Reply Last reply Reply Quote 3
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