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QRect and wrong rectangle coordinates

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  • J Offline
    J Offline
    JulienD
    wrote on last edited by JulienD
    #1

    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

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    • B Offline
      B Offline
      Bonnie
      wrote on last edited by Bonnie
      #2

      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)

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      • J Offline
        J Offline
        JulienD
        wrote on last edited by
        #3

        Missed that info. Everything is alright. thanks!

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        0
        • HoMaH Offline
          HoMaH Offline
          HoMa
          wrote on last edited by
          #4

          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)
          
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          3

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