Problem with a color class

[zorro68]

Yes, this works fine if there is a simple function. But now I'm going to use the "colors" variable in another function and i have an error because this function cannot "see" this variable.
So I put in my .h file the definition of the "colors" variable like this: QList<QColor *> colors; (¿is this ok?)
and I'm going to initialize theirs values in .cpp file with this
colors << QColor(0,0,0) << QColor(255,255,255) << ...
but this not work because the << operator don't work with pointers (I think).
How can i fill the values of colors?

[wysota]

Yes, this works fine if there is a simple function. But now I'm going to use the "colors" variable in another function and i have an error because this function cannot "see" this variable.

So make it global to the parent scope (for example by making it a member variable) or global to the application or pass it as an argument to where you want to use it.

So I put in my .h file the definition of the "colors" variable like this: QList<QColor *> colors; (¿is this ok?)

No. Should be "extern QList<QColor> colors;" (and forget the pointer, there is no point using it and you're wasting additionally 4/8 bytes per item). Then create the variable in one of your .cpp files without the extern keyword.

and I'm going to initialize theirs values in .cpp file with this
colors << QColor(0,0,0) << QColor(255,255,255) << ...
but this not work because the << operator don't work with pointers (I think).
How can i fill the values of colors?

Don't use pointers

[zorro68]

Should be "extern QList<QColor> colors;" . Then create the variable in one of your .cpp files without the extern keyword.

So you say that i have to write "extern QList<QColor> colors;" in my .h file out of the class (because .h file conteins a class) and "QList<QColor> colors"; in the function where i want to use this variable in the .cpp file.??

Note that i have defined this variable in the constructor of a class and using in other functions of the class.

[zorro68]

I have a lots of troubles with the extern variable and to avoid this extern variable, i have try this:

foreach(QColor color, colors){
	CMBcolor[i]->addItem(QString::null,color);
	int index = CMBcolor[i]->count()-1;
	pixmap.fill(color);
	CMBcolor[i]->setItemData(index, pixmap, Qt::DecorationRole);
}

foreach(QColor color, colors){
	CMBcolor[i]->addItem(QString::null,color);
	int index = CMBcolor[i]->count()-1;
	pixmap.fill(color);
	CMBcolor[i]->setItemData(index, pixmap, Qt::DecorationRole);
}

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and fill all my combos correctly (I think this is more elegant). But when I want to know the clicked color i program (in other functions):

QColor col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

QColor col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

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(ncolor is the currentitem in the combo, sup is the combo i have picked)

But I obtain an error. QColor is not QVariant. I have proved this:

QVariant col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

QVariant col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

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and compile ok but how can i obtain the r chanel, g chanel and b chanel of the color from col.

Note that i would also have this problem with the other way too.

[wysota]

So you say that i have to write "extern QList<QColor> colors;" in my .h file out of the class (because .h file conteins a class) and "QList<QColor> colors"; in the function where i want to use this variable in the .cpp file.??

No, not in a function. Outside all functions - this will make it global to the application. And the extern statement just says "hey you compiler! There is a 'QList<QColor> colors' variable somewhere out there so you can allow functions to use it, it will be available during link time".

Note that i have defined this variable in the constructor of a class and using in other functions of the class.

You should define it in main.cpp (outside main()) and fill it in main().

I have a lots of troubles with the extern variable and to avoid this extern variable, i have try this:

foreach(QColor color, colors){
	CMBcolor[i]->addItem(QString::null,color);
	int index = CMBcolor[i]->count()-1;
	pixmap.fill(color);
	CMBcolor[i]->setItemData(index, pixmap, Qt::DecorationRole);
}

foreach(QColor color, colors){
	CMBcolor[i]->addItem(QString::null,color);
	int index = CMBcolor[i]->count()-1;
	pixmap.fill(color);
	CMBcolor[i]->setItemData(index, pixmap, Qt::DecorationRole);
}

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and fill all my combos correctly (I think this is more elegant).

Not really. It would be better to subclass QComboBox and do that inside its constructor. And of course then use the subclass instead of QComboBox.

But I obtain an error. QColor is not QVariant. I have proved this:

QVariant col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

QVariant col = CMBcolor[sup]->itemData(ncolor,Qt::UserRole);

To copy to clipboard, switch view to plain text mode 

Of course it's not. You have to convert it to QColor. I suggest you read QVariant docs, especially the section about qvariant_cast.

Maybe you should just use QwwColorComboBox?

[zorro68]

At last, the combo works fine.

Wysota thank you very much for your help and for your patience. Think that i am newbie in qt and c++ so i don't know a lot of "easy" ways to do something in qt.

I'm programming an scientific aplication in fortran (this works fine) and I need a grafical interface to introduce data, but to work with windows and linux, so this is why i use qt. I have program in fortran, c, visual basic, ... and i thought that c++ would be like c, but it isn't. Now i have to plot results (data surface) in 3d, so i'm programming an opengl module (marching cubes algorithm) to do this. ¡¡¡¡¡¡¡ C++, qt, opengl, math algorithms all in one, buffff !!!!!!!! and english. In two weeks, i'll go to uk (Cardiff) with a grant to spend 10 month (I hope to learn a good english).

I'll be here soon with other question...

Now a personal question for you and jpn (you can answer if you like), have you programm the qt or do you work for qt? (because you are connected all time as i can see).

Thanks for all

[wysota]

We don't work for Trolltech if that is what you ask.