Qt 5.3 Signals and Slots, Simple Function and Lambda Expression

Question

I tried to Write a program using Qt 5.3 and I try to using signals and slots as practice. I wrote the following code (part of the code) :

void exitToWin()
    {
        exit(0);
    }
int main(int argc, char *argv[])
{

    QApplication a(argc, argv);
    MainWindow w;
    QMessageBox EndBox;
    QObject::connect((EndBox.button(QMessageBox::Ok)),SIGNAL(clicked()),exitToWin);
    w.show();
    EndBox.show();
    return a.exec();
}

I even change the declaration of the function to static and I checked the expression with parentheses and without them while I am writing the connect command. but although what Qt documented and what its IDE guided to. also I read here and I tested it.
Moreover I tried with lambda expression as below:

QObject::connect((EndBox.button(QMessageBox::Ok)),SIGNAL(clicked()),[=](){
    exit(0);
});

but still I receive errors indicate "No matching function call".
And after all I have to say that I am using Microsoft Windows 7.

Did youu try to rewrite that wihout SIGNAL declaration like here: http://woboq.com/blog/cpp11-in-qt5.html ? — Tanuki, Aug 13 '14 at 06:43

score 4 · Accepted Answer · edited Aug 13 '14 at 08:12

4

This works on Qt 5.3:

#include <QtWidgets>

void exitToWin()
{
    exit(0);
}

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);
    QMainWindow w;
    QMessageBox endBox;
    endBox.addButton(QMessageBox::Ok); // (2)
    endBox.connect(endBox.button(QMessageBox::Ok), 
                   &QAbstractButton::clicked, exitToWin); // (1)
    /* This works, too:
    endBox.connect(endBox.button(QMessageBox::Ok),
                   &QAbstractButton::clicked,
                   [] () { exit(0); });
    */
    w.show();
    endBox.show();
    return a.exec();
}

Here's why:

(1) You can use endBox's QObject to do the connection between the QAbstractButton's clicked signal and your exitToWin simple function. You also can't connect a SIGNAL to a simple function or a lambda, so we use the member function variety, instead.

(2) endBox doesn't actually get an OK button by default. When you mention it on line (1) in your code, it creates it, but not in time (apparently) to pass the pointer back to connect, so we create it first here.

edited Aug 13 '14 at 08:12

Martin Delille

11,360
15
65
132

answered Aug 13 '14 at 06:43

Alex Reinking

16,724
5
52
86

Thank you very much, I tested something else, as you mentioned in line 2 we can use `QPushButtom` too, it work, I think the problem relates to what you mentioned in line 2. – bahrami307 Aug 13 '14 at 06:52
the Problem is still with lambda expression, can you help? – bahrami307 Aug 13 '14 at 06:56
make sure you have the `-std=c++11` flag set. – Alex Reinking Aug 13 '14 at 06:59
I added in Built steps but the I received the unknown command error – bahrami307 Aug 13 '14 at 07:07

score 0 · Answer 2 · answered Aug 13 '14 at 06:41

Your code won't work for two reasons:

Firstly, QMessageBox does not have such a signal. See the documentation for the signals it does have.

Secondly, when making connections from a signal to a slot (or lambda function), you must define the function signatures, not specific values.

If a signal can pass a variety of values and you only want your slot to perform a certain function on a selection of those values (in this case, only if the value QMessageBox::Ok is passed) it is up to the slot to interrogate the values, not the connect statement.

score -1 · Answer 3 · answered Aug 13 '14 at 06:36

-1

Since the connect() method is from QObject it has to be called from a QObject child containing the Q_OBJECT macro in its declaration. Running qmake prepare the class to send signals and receive slots.

answered Aug 13 '14 at 06:36

Martin Delille

11,360
15
65
132

`QObject::connect` is a static function. – RobbieE Aug 13 '14 at 06:41

Qt 5.3 Signals and Slots, Simple Function and Lambda Expression

3 Answers3