invalid conversion from 'QWidget*' to 'QwtPlotCanvas*' [-fpermissive]

Question

I'm trying to compile a program (found here: http://sourceforge.net/projects/lisem/) by following the instructions said by the author. However, when compiling it in Qt Creator, it gives the error:

invalid conversion from 'QWidget*' to 'QwtPlotCanvas*' [-fpermissive]

for this line in LisUImapplot.cpp

186 picker = new MyPicker( MPlot->canvas() );

link to its header file (LisUImapplot.h) can be found on the same folder as the cpp file.

class MyPicker: public QwtPlotPicker
{
public:
    MyPicker( QwtPlotCanvas *canvas ):
        QwtPlotPicker( canvas )
    {
        setTrackerMode( AlwaysOn );
    }

    virtual QwtText trackerTextF( const QPointF &pos ) const
    {
        QColor bg( Qt::white );
        bg.setAlpha( 100 );

        QwtPlotItemList list = plot()->itemList(QwtPlotItem::Rtti_PlotSpectrogram);
        QwtPlotSpectrogram * sp = static_cast<QwtPlotSpectrogram *> (list.at(1));
        double z = sp->data()->value(pos.x(), pos.y());
        QString txt = "";
        if (z > -1e10)
            txt.sprintf( "%.3f", z );
        QwtText text = QwtText(txt);
        text.setColor(Qt::black);
        text.setBackgroundBrush( QBrush( bg ) );
        return text;
    }
};

I hope you can help me on this. Thank you!

I am using Qt 5.1.1 MinGW 32-bit and Qwt 6.1.0

Answer 1

QwtPlot::canvas() returns a QWidget. Your MyPicker constructor is expecting a QwtPlotCanvas type parameter.

You can cast it to a QwtPlotCanvas:

QwtPlotCanvas *canvas = qobject_cast<QwtPlotCanvas*>(MPlot->canvas());
if(canvas)
{
   picker = new MyPicker(canvas);
   ...
}

Answer 2

Why do you do constructor?

MyPicker( QwtPlotCanvas *canvas ):
    QwtPlotPicker( canvas ){}  

In the old version of Qwt was QwtPlotPicker::QwtPlotPicker(QwtPlotCanvas canvas); In the Qwt 6.1. is QwtPlotPicker::QwtPlotPicker(QWidget *parent);

You have to do

MyPicker( QWidget *canvas ):
    QwtPlotPicker( canvas ){}