Error : "Cannot take the address of an rvalue of type 'void' "

[MortyMars]

Hello everyone,

Thanks to the help of the forum, I'm making progress with my project, but a new problem has arisen with this error: "cannot take the address of an rvalue of type 'void'".

The error is there:

The method called - which compiles without a problem - is this one:

So far (but my experience is almost non-existent) I've never encountered this problem when 'connecting' and I have the impression that the blocking comes from the call to a method requiring parameters...

Thank you for your suggestions

[JonB]

@MortyMars
Please post code text not images. I have to type in what the image shows now.

You cannot have &FixWiindow::CalculFixType(0,1) for a "PointerToMemberFunction method". It must be a (pointer to a) method like &FixWiindow::CalculFixType, no parentheses. Normally a slot takes any parameters passed by the signal. If you intend to call that function with some parameters of your own use a lambda instead.

[MortyMars]

@JonB

Thank, and sorry for the extra work :-(

A little late, but the offending code is here:

connect(ui->cboWP0_1, &QComboBox::textActivated,
       this, &FixWindow::CalculFixType(0,1) );

And the method called, here :

// VERSION 2 ADAPTEE A TOUTES LES TRIPLETTES DE COMBOBOX
void FixWindow::CalculFixType(int col, int ligne)
{
    // Recup des codes ASCII de chaque lettre des 3 combobox
    // Les codes, à ce stade, sont en base 10
    int ASC1 = ComboMatrix.at(col).at(ligne)->currentText().left(1).at(0).toLatin1();
    int ASC2 = ComboMatrix.at(col+1).at(ligne)->currentText().left(1).at(0).toLatin1();
    int ASC3 = ComboMatrix.at(col+2).at(ligne)->currentText().left(1).at(0).toLatin1();

    // Conversion des codes en base 16 et inversion de l'ordre
    // pour constituer la chaine hexa à reconvertir en base 10
    QString HexaStr = QString("%1").arg(ASC3, 2, 16, QLatin1Char('0')) +
                      QString("%1").arg(ASC2, 2, 16, QLatin1Char('0')) +
                      QString("%1").arg(ASC1, 2, 16, QLatin1Char('0')) ;

    // Conversion de la chaine en entier en base 10
    bool ok;
    int DeciNum = HexaStr.toInt(&ok,16);


    TxtMatrix.at(ligne)->setText(QString::number(DeciNum));

}

What exactly do you mean by a 'lambda'?

[ChrisW67]

This is what your connect should look like:

connect(ui->cboWP0_1, &QComboBox::textActivated, this, &FixWindow::CalculFixType);

except this will also fail. QComboBox::textActivated() is going to pass a QString value to any slot you connect it to. Typically that slot needs to accept the string as a parameter. Your proposed slot expects two integer arguments.

In this context, a C++ lambda expression is an unnamed callable code block that can accept/capture parameters and act as the target of a connect (in place of &FixWindow::CalculFixType). Something roughly like this (unable to test at the moment):

connect(
  ui->cboWP0_1, &QComboBox::textActivated, 
  this, [&, c = 0, l = 1](const QString &text) {
    this->CalculFixType(c, l);
  }
);

I am sure some more frequent lambda user will correct this.

If you are doing this for a matrix of combo boxes then there are other ways you could achieve this.

[MortyMars]

@ChrisW67

Great !!! Thanks Chris,
With a little adaptation of the code (see below), it works perfectly:

connect(ui->cboWP0_1, &QComboBox::textActivated,
            this, [&, col=0, ligne=1]{FixWindow::CalculFixType(col,ligne);} );

I had an "expected body of lambda expression" error, corrected by adding {...}

And you're right, this method uses a combobox matrix (calculation of the numerical code of an "Enroute and Terminal Waypoint")

[JonB]

@MortyMars
You would normally write this->CalculFixType(...) or just CalculFixType(...) where you have FixWindow::CalculFixType(...) in the lambda's body.

[MortyMars]

OK, thank you @JonB for that clarification.

I'm still having a bit of trouble understanding what 'this' means in different contexts...

[jsulm]

@MortyMars "this" is always the pointer to the current class instance

[JonB]

@MortyMars
To do Object-Oriented/C++ programing you need to understand class vs instance. There will be myriad explanations on the web, so I'm not going to try to explain :)

I will say: in code for every one time you need to write ClassName::methodName somewhere there are 99 cases where you want this->methodName (or just plain methodName, you don't need to insert the this-> before it)!

[MortyMars]

@jsulm said in Error : "Cannot take the address of an rvalue of type 'void' ":

@MortyMars "this" is always the pointer to the current class instance

Thank you for this clear and precise reminder ;-)

[MortyMars]

@JonB
Far be it from me to ask you for basic lessons (sorry for my often naive questions) but I've only recently started learning C++ again and a lot of reflexes have been lost.
So thank you in advance for your patience, I'm improving (I think) a little more every day ;-)
And thank you of course for all your tips and advice :-)