no match for 'operator>>' (operand types are 'QDataStream' and 'double')

[Cruz]

Hello!

I would like to be able to serialize my custom vector class to a QDataStream, and for some reason that I don't understand, my operator overloading doesn't work.
Here is the relevant part of my vector class:

#include <QDataStream>
 
template <int N=3>
class NVec
{
    double v[N];
 
public:
 
    NVec() {memset(v, 0, N*sizeof(double));}
};
 
template <int N>
QDataStream& operator<<(QDataStream& out, const NVec<N> &o)
{
    for (int i = 0; i < N; i++)
        out << o[i];
    return out;
}
 
template <int N>
QDataStream& operator>>(QDataStream& in, const NVec<N> &o)
{
    for (int i = 0; i < N; i++)
        in >> o[i];
    return in;
}

#include <QDataStream>

template <int N=3>
class NVec
{
    double v[N];

public:

    NVec() {memset(v, 0, N*sizeof(double));}
};

template <int N>
QDataStream& operator<<(QDataStream& out, const NVec<N> &o)
{
    for (int i = 0; i < N; i++)
        out << o[i];
    return out;
}

template <int N>
QDataStream& operator>>(QDataStream& in, const NVec<N> &o)
{
    for (int i = 0; i < N; i++)
        in >> o[i];
    return in;
}

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The compiler complains with

no match for 'operator>>' (operand types are 'QDataStream' and 'double')

even though I am certain that I have used QDataStream with doubles before. It is also strange that only the read operator fails, but not the write operator.
Can anyone please point out where I went wrong?

Thanks
Cruz

[d_stranz]

For operator>>(), the NVec<N> argument has to be non-const. (You're modifying it after all).

Cut-n-paste strikes again, eh?

[Cruz]

Oh crap. I am the stupidest stupid in the world.

[d_stranz]

Nope, I've done it too. I was stupid before you were.

[Cruz]

You _were_, but now I have to wait for someone else to make that mistake in order to pass on the stupid.

[simex]

Yeap, I've come to collect the stupid Feel free to pass it on
I have a similar problem here except the problem persists even after the argument was changed to a non-const. Any idea as to why that might happen? Below is the implementation of the two operators.

QDataStream &operator <<(QDataStream &out, const PID &pid)
{
    out << pid.getKp() << pid.getKi() << pid.getKd();
    return out;
}
 
 
QDataStream &operator >>(QDataStream &in, PID &pid)
{
    in >> pid.getKp() >> pid.getKi() >> pid.getKd();
    return in;
}

QDataStream &operator <<(QDataStream &out, const PID &pid)
{
    out << pid.getKp() << pid.getKi() << pid.getKd();
    return out;
}


QDataStream &operator >>(QDataStream &in, PID &pid)
{
    in >> pid.getKp() >> pid.getKi() >> pid.getKd();
    return in;
}

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[anda_skoa]

Do the getter functions, e.g. PID.getKp(), return a reference to which the >> operator could write?

Cheers,
_

[simex]

Yes that was the problem. Thank you. I changed it to the code below and now it's okay. I guess I'll wear my cap for while Thanks again.

QDataStream &operator >>(QDataStream &in, PID &pid)
{
    double kp;
    double ki;
    double kd;
    in >> kp >> ki >> kd;
    pid.setKp(kp);
    pid.setKi(ki);
    pid.setKd(kd);
    return in;
}

QDataStream &operator >>(QDataStream &in, PID &pid)
{
    double kp;
    double ki;
    double kd;
    in >> kp >> ki >> kd;
    pid.setKp(kp);
    pid.setKi(ki);
    pid.setKd(kd);
    return in;
}

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