[SOLVED] QLabel won't update pixmap from inside function

SGaist

Can you try with the following widget:

widget.h

#ifndef WIDGET_H
#define WIDGET_H

#include <QWidget>

class QLabel;

class Widget : public QWidget
{
    Q_OBJECT

public:
    Widget(QWidget *parent = 0);
    ~Widget();

public slots:
    void onClicked();

private:
    QVector<QLabel*> _labels;
    int _currentLabel;
    QColor _color;
};

#endif // WIDGET_H

widget.cpp

#include "widget.h"
#include <QString>
#include <QLabel>
#include <QPushButton>
#include <QHBoxLayout>
#include <QVBoxLayout>

Widget::Widget(QWidget *parent)
    : QWidget(parent)
    , _currentLabel(0)
    , _color(Qt::blue)
{
    QHBoxLayout *buttonLayout = new QHBoxLayout;
    QImage img(120, 120, QImage::Format_ARGB32);
    img.fill(Qt::red);
    for (int i = 0 ; i < 4 ; ++i) {
        QLabel *label = new QLabel;
        label->setPixmap(QPixmap::fromImage(img));
        buttonLayout->addWidget(label);
        _labels << label;
    }

    QPushButton *button = new QPushButton(tr("Test"));

    QVBoxLayout *layout = new QVBoxLayout(this);
    layout->addLayout(buttonLayout);
    layout->addWidget(button);

    connect(button, &QPushButton::clicked, this, &Widget::onClicked);
}

Widget::~Widget()
{

}

void Widget::onClicked()
{
    QImage img(120, 120, QImage::Format_ARGB32);
    img.fill(_color);

    _labels[_currentLabel]->setPixmap(QPixmap::fromImage(img));
    ++_currentLabel;
    if (_currentLabel == _labels.count()) {
        _currentLabel = 0;
        _color = QColor(qrand() % 255, qrand() % 255, qrand() % 255);
    }
}

main.cpp

#include "widget.h"
#include <QApplication>

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    Widget w;
    w.show();

    return a.exec();
}

Tymer

That's working perfectly for me. Maybe this will give me the clue I need to make it work...

Tymer

Ah, that one works because it is from within the same widget. I added a push button to myWidget and it was working. Unfortunately, the button that I need to push is in a different window.

~~ EDIT: The new button on myWidget works perfectly. I thought maybe I could circumvent the problems of the changeImage button by making it a slot for "emit buttonclicked," which I connected to the new button on myWidget. It had the same problem as before, where it does nothing and prints "end of array" after 11 clicks. BUT strangely, if I press the button on myWidget, it resets the value of activePointer. I'm not sure why this is.

Essentially, I can push the changeImage button 11 times until it prints "end of array," but then I can push the myWidget pushbutton 11 times and it will change the image 10 times and start printing "end of array" after that. Curious. ~~

EDIT EDIT: It looks like right now it's only changing the value of activePointer locally. Whoops. Problem still exists, however.

SGaist

The fact that the button is inside another widget should not have any impact here. Can you reproduce your bug using may sample as base ?

Tymer

I'll post the full code, just to be thorough. The error is exactly the same. When I debugged, it showed that my signal had reached the slot...But there was no change to the image.

widget1.h

#ifndef WIDGET1_H
#define WIDGET1_H

#include <QWidget>
#include "/absolute/path/to/widget.h"

namespace Ui {
class widget1;
}

class widget1 : public QWidget
{
    Q_OBJECT

    Widget* widget;

public:
    explicit widget1(QWidget *parent = 0);
    ~widget1();

signals:
    void clicked();
    void buttonPush();

private slots:
    void on_pushButton_clicked();

private:
    Ui::widget1 *ui;
};

#endif // WIDGET1_H

widget1.cpp

#include "widget1.h"
#include "ui_widget1.h"

widget1::widget1(QWidget *parent) :
    QWidget(parent),
    ui(new Ui::widget1)
{
    ui->setupUi(this);

    widget = new Widget;

    connect(this, SIGNAL(buttonPush()),widget,SLOT(onClicked()));
}

widget1::~widget1()
{
    delete ui;
}

void widget1::on_pushButton_clicked()
{
    emit buttonPush();
}

An instance of widget1 was created in main, then shown. I created a ui file in designer, with all default values.

SGaist

Is it me or are you not showing widget at all ?

Tymer

My main.cpp:

#include "widget.h"
#include "widget1.h"
#include <QApplication>

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    Widget w;
    widget1 w1;
    w1.show();
    w.show();


    return a.exec();
}

SGaist

Ok, you are not showing the widget you have connected. Your Widget w in your main.cpp is not the same as the one you have instantiated in your widgte1 constructor.

Tymer

So I should make my widget a singleton, more or less? How would I go about that?

SGaist

No, not at all. Either:

Remove the widget from MainWindow and connect the one in main.cpp
Remove the one from main.cpp and show the one from MainWindow

Tymer

How would I connect to the instance in main?

SGaist

By using the static version of QObject::connect

Tymer

I know this is really basic, but I'm having a bit of a brain fart. I want to make a pointer to the instance I create in main, but I don't want my class to know too much about what's going on in main. How do I do this so that I can create a connection to the object in main from a different widget?

Tymer

Right now I have:

main.cpp


myWidget* w = new myWidget;
actionTestWindow* t = new actionTestWindow(w, new QWidget);

actionTestWindow.cpp

actionTestWindow::actionTestWindow(myWidget* Widget, QWidget *parent) :
    QWidget(parent),
    ui(new Ui::actionTestWindow)
    {
        ui->setupUi(this);

        Widget = new myWidget;

The actionTestWindow is no longer showing, and I think it has something to do with how I insert the QWidget parameter.

SGaist

Roll back to before you modified that constructor. Just call '''Widget->show();'''

Tymer

Alright, finally got this problem SOLVED!

I pass in an instance of myWidget as a parameter of actionTestWindow in main. It's important to note that you need to specify the QWidget as parent = 0 or it will not show. The relevant code looks something like this:

main.cpp

QWidget* parent = new QWidget;
parent = 0;
myWidget* w = new myWidget;
actionTestWindow* a = new actionTestWindow(w,parent);

w->show();
a->show();

actionTestWindow.h

class actionTestWindow : public QWidget {
myWidget* m;

public:
actionTestWindow(myWidget* mW, QWidget* parent = 0);

actionTestWindow.cpp

actionTestWindow::actionTestWindow(myWidget* mW, QWidget* parent)
{
this->m = mW;
/*use this->m to reference myWidget*/
}

**note: irrelevant or standard code was mostly emitted.

SGaist

And there you have a memory leak.

Why don't you just show the myWidget you had originally created in Widget ?

Tymer

@SGaist It's a pretty complicated project. The actionTestWindow is pretty much just what it sounds like...A test window. The point is to make sure that I can successfully pass the signals I need and get the reactions I need for when I connect it to a more complicated process. myWidget needs to be shown from main as soon as the program starts running. Having it wait to show until the process that it displays info from runs would be a problem. The code will be passed around to many other people to be changed for future projects, and we're trying to avoid making them dig through code to find things as much as possible. It needs to be easy to just change a class slightly and change the instantiation slightly in order to make a change.

Can you point out the exact memory leak? I was under the impression that you can pass in instances of classes as parameters. It's a pretty common thing to do in Java, and I've at least heard of it being done in C++. There is a lot of code omitted. Destructors, importantly.

SGaist

You have

QWidget* parent = new QWidget;
parent = 0; << the original parent is now lost and not destroyed
myWidget* w = new myWidget;

Since it needs to be the first widget, then create it in main. Connect your MainWindow to it also in main.

You should rather avoid making that test widget known to any and every widget of your software. Doing so you'll avoid tight coupling and maintenance hell.

Tymer

Changed it to QWidget* parent = 0;. No more memory leak, everything works as expected.