[PyQt] SIP gives error when trying to subclass QVariant
Nate Reid
gnatty7 at hotmail.com
Tue Apr 16 19:53:37 BST 2013
Versions: Redhat Linux EL6Qt 4.7.1 (also tested with 4.8.2)PyQt: 4.10Python: 2.7.4Sip: 4.14
Sip 4.14, and 4.13 as well gives me the following error when trying to generate code on the following sip file:sip -x VendorID -t WS_X11 -x PyQt_NoPrintRangeBug -t Qt_4_7_1 -g -I /usr/pic1/root/share/sip/PyQt4 -c . python/TestQVariant.sipsip: python/TestQVariant.sip:6: A class, exception, namespace or mapped type has already been defined with the same name
// TestQVariant.sip:%Module TestQVariant
%Import QtCore/QtCoremod.sip
class TestQVariant : public QVariant{%TypeHeaderCodeinclude "TestQVariant.h"$Endpublic: TestQVariant(void);
TestQVariant(const QVariant &v); TestQVariant(int v); TestQVariant(double v); TestQVariant(const QString &v);
TestQVariant(const TestQVariant &a);
bool hasName(void) const; const QString &name(void) const; void setName(const QString &name);
};%End
// TestQVariant.h#ifndef TestQVariant_h#define TestQVariant_h
#include <QtCore/QVariant>#include <QtCore/QString>
class TestQVariant : public QVariant{public: TestQVariant(void) { }
TestQVariant(const QVariant &v) : QVariant(v) { } TestQVariant(int v) : QVariant(v) { } TestQVariant(double v) : QVariant(v) { } TestQVariant(const QString &v) : QVariant(v) { }
TestQVariant(const TestQVariant &a) : QVariant(a), mName(a.mName) { }
TestQVariant &operator=(const TestQVariant &a) { QVariant::operator=(a); mName = a.mName; return *this; } bool hasName(void) const { return !mName.isEmpty(); } const QString &name(void) const { return mName; } void setName(const QString &name) { mName = name; } private: QString mName;};
#endif
Is this related to the API version that QVariant is using? This is with Python 2.7, so it should default to '2'. Is there some other name I need to use for the QVariant class to get this to work? -Nate
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