在QT中重载QDebug

Question

In view.h file:

friend QDebug operator<< (QDebug , const Model_Personal_Info &);

view.cpp文件:

QDebug operator<< (QDebug out, const Model_Personal_Info &personalInfo) {
    out << "Personal Info :n";
    return out;
}

调用后:

qDebug() << personalInfo;

假设给出输出:"Personal Info :"

但是给出错误:

error: no match for 'operator<<' in 'qDebug()() << personalInfo'

Answer 1

Header:

class DebugClass : public QObject
{
    Q_OBJECT
public:
    explicit DebugClass(QObject *parent = 0);
    int x;
};
QDebug operator<< (QDebug , const DebugClass &);

和实现:

DebugClass::DebugClass(QObject *parent) : QObject(parent)
{
    x = 5;
}   
QDebug operator<<(QDebug dbg, const DebugClass &info)
{
    dbg.nospace() << "This is x: " << info.x;
    return dbg.maybeSpace();
}

或者你可以这样定义all in header:

class DebugClass : public QObject
{
    Q_OBJECT
public:
    explicit DebugClass(QObject *parent = 0);
    friend QDebug operator<< (QDebug dbg, const DebugClass &info){
        dbg.nospace() << "This is x: " <<info.x;
        return dbg.maybeSpace();
    }
private:
    int x;
};

Answer 2

尽管当前的答案可以解决问题，但是其中有很多多余的代码。把这个加到你的.h里。

QDebug operator <<(QDebug debug, const ObjectClassName& object);

然后在你的.cpp中这样执行。

QDebug operator <<(QDebug debug, const ObjectClassName& object)
{
    // Any stuff you want done to the debug stream happens here.
    return debug;
}