CODEWITHSUNDEEP
c++qt4operator-overloading
izidor
izidor

Reputation: 4144

How to overload operator<< for qDebug

I'm trying to create more useful debug messages for my class where store data. My code is looking something like this

#include <QAbstractTableModel>
#include <QDebug>

/**
  * Model for storing data. 
  */
class DataModel : public QAbstractTableModel {
    // for debugging purposes
    friend QDebug operator<< (QDebug d, const DataModel &model);

    //other stuff
};

/**
  * Overloading operator for debugging purposes
  */
QDebug operator<< (QDebug d, const DataModel &model) {
    d << "Hello world!";
    return d;
}

I expect qDebug() << model will print "Hello world!". However, there is alway something like "QAbstractTableModel(0x1c7e520)" on the output.

Do you have any idea what's wrong?

Upvotes: 28

Views: 24581

Answers (5)

fferri
fferri

Reputation: 18950

You implemented only the << operator for a reference. If your model variable is a pointer, it will use another implementation (not yours).

To use your implementation you can do:

qDebug() << *model

By the way, the correct way to implement a QDebug operator<<(QDebug dbg, const T &data) overload is to use the QDebugStateSaver class:

QDebug operator<<(QDebug dbg, const QDataflowModelOutlet &outlet)
{
    QDebugStateSaver stateSaver(dbg);
    dbg.nospace() << ...;
    return dbg;
}

In this way the settings (i.e. wether to insert or not spaces between prints) will be correctly restored when exiting the function.

Upvotes: 10

Eliyahu Skoczylas
Eliyahu Skoczylas

Reputation: 359

I found this answer on the QT Forum by raven-worx (giving credit where credit is due!)

In the .h file:

QDebug operator<<(QDebug dbg, const MyType &type);

where MyType is your class, like DataModel and type is the instance you will display.

And in the .cpp file:

QDebug operator<<(QDebug dbg, const MyType &type)
{
    dbg.nospace() << "MyType(" << .... << ")";
    return dbg.maybeSpace();
}

and you can use the QDebug's space(), nospace(), and other methods to control the exact display of the stream.

So for the OP, we would use:

// in the .h file:
class DataModel : public QAbstractTableModel {
// stuff
};
QDebug operator<<(QDebug dbg, const DataModel &data);

// in the .cpp file:
QDebug operator<<(QDebug dbg, const DataModel &data)
{
    dbg.nospace() << "My data {" << data.someField << ',' << data.another << "}";
    return dbg.maybeSpace();
}

// in some .cpp user of the class:
DataModel myData;

. . .

QDebug() << "The current value of myData is" << myData;

Upvotes: 3

Vin&#237;cius A. Jorge
Vin&#237;cius A. Jorge

Reputation: 738

I know it long time now, but just to be documented and to help any other people who eventually come here having the same doubt, the easiest way to get qDebug() << working with your own class printing something like "Hello World" or whatever else, is to implement implicit conversion of your class to a printable type, like QString (which is well supported by QDebug).

class Foo {
public:
   Foo() { }
   operator QString() const { return <put your QString here>; }   

};

Upvotes: 34

a.l.e
a.l.e

Reputation: 714

In your example, qDebug() prints the address of your variable, which is the default behavior for unknown types.

In fact, there seem to be two things you have to take care of:

  • Get the item by value (and eugen already pointed it out!).
  • Define the overloading operator before you use it, put its signature in the header file, or define it as forward before using it (otherwise you will get the default "qDebug() <<" behavior).

This will give you:

QDebug operator<< (QDebug d, const DataModel &model) {
    d << "Hello world!";
    return d;
}
DataModel m;
qDebug() << "m" << m;

or

QDebug operator<< (QDebug d, const DataModel &model);

DataModel m;
qDebug() << "m" << m;

QDebug operator<< (QDebug d, const DataModel &model) {
    d << "Hello world!";
    return d;
}

I've learned it the hard way, too...

Upvotes: 15

izidor
izidor

Reputation: 4144

After an hour of playing with this question I figured out model is pointer to DataModel and my operator << takes only references.

Upvotes: 12

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