Var in Lambda pass out ?

sonichy

error: assignment of read-only variable ‘bv’ : bv = v;

void MainWindow::on_action_brightness_triggered()
{
    int bv = 0;
    QDialog *dialog = new QDialog;
    dialog->setWindowTitle("Brightness");
    dialog->setFixedSize(400,200);
    QSlider *slider = new QSlider(Qt::Horizontal);
    slider->setRange(-50, 50);
    slider->setTickPosition(QSlider::TicksBelow);
    connect(slider, &QSlider::valueChanged, [=](int v){
        bv = v;
        imageWidget->adjustBrightness(v, false);
    });
    QVBoxLayout *vbox = new QVBoxLayout;
    vbox->addWidget(slider);
    QPushButton *pushButton_confirm = new QPushButton("Confirm");
    QPushButton *pushButton_cancel = new QPushButton("Cancel");
    QHBoxLayout *hbox = new QHBoxLayout;
    hbox->addStretch();
    hbox->addWidget(pushButton_confirm);
    hbox->addWidget(pushButton_cancel);
    hbox->addStretch();
    vbox->addLayout(hbox);
    dialog->setLayout(vbox);
    connect(pushButton_confirm, SIGNAL(clicked()), dialog, SLOT(accept()));
    connect(pushButton_cancel, SIGNAL(clicked()), dialog, SLOT(reject()));
    if(dialog->exec() == QDialog::Accepted){
        imageWidget->adjustBrightness(bv, true);
    }
}

sierdzio

bv goes out of scope when on_action_brightness_triggered() method ends (}). Your lambda is called afterwards, when the variable does not exist anymore.

Make bv a class member, not local variable and it won't show an error.

sonichy

My net friend show me the simplest way : change lambda [=] to [&].
It works, but I do not know why.

JKSH

@sierdzio said in Var in Lambda pass out ?:

bv goes out of scope when on_action_brightness_triggered() method ends (}). Your lambda is called afterwards, when the variable does not exist anymore.

In this case, dialog->exec() blocks and runs a new event loop so bv technically remains in scope until the dialog and the connection gets destroyed.

It's a fragile design, however.

@sonichy said in Var in Lambda pass out ?:

My net friend show me the simplest way : change lambda [=] to [&].
It works, but I do not know why.

[=] gives your lambda a copy of bv.
[&] gives your lambda a reference to bv.

See the answer of https://stackoverflow.com/questions/7627098/what-is-a-lambda-expression-in-c11