[PyQt] SIP fails with "syntax error" on simple class inheritance
tuxor1337 at web.de
tuxor1337 at web.de
Mon Apr 30 11:27:03 BST 2012
Why does the attached sip-file always fail to be "sipped" with the
following error message:
sip: my_dialog.sip:15: syntax error
(line 15 is "class MyDialog : public QDialog {")
I tried to use a configure.py or directly call sip command line, but it
always fails with the same error message.
What am I doing wrong? Is there a tutorial out there that's a bit easier
to understand than the official documentation
(http://www.riverbankcomputing.co.uk/static/Docs/sip4/using.html)? The
documentation is rather useless when using SIP for bigger or more
complex libraries.
###### Code of my_dialog.sip #######################
%Module myDialog
%Import QtGui/QtGuimod.sip
%If (Qt_4_2_0 -)
namespace ui {
%TypeHeaderCode
#include "my/ui/buttons.h"
#include "my/ui/label.h"
#include "my/ui/soft_shadow.h"
%End
class MyDialog : public QDialog {
public:
MyDialog(QWidget *parent, bool show_shadow = true);
~MyDialog(void);
void updateTitle(const QString &message);
public Q_SLOTS:
virtual void done(int);
};
%End
};
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