How to determine type of widget in a qtable cell?

Question

I have created a QTable with lots of gui elements like comboBoxes and checkBoxes in various cells. I am able to access these elements by creating pointers to them. What I want to know is, is there any way to know what type of widget(comboBox or checkBox) a cell is having?

score 16 · Accepted Answer · edited Feb 04 '19 at 06:48

16

Check out the answers to this question. The accepted answer gets the class name (as a const char*) from the widget's meta-object like so:

widget->metaObject()->className();

There's another answer that suggests using C++'s type management, but that sounds a lot less wieldly (more unwieldy?).

edited Feb 04 '19 at 06:48

AAEM

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answered Jun 01 '11 at 18:33

Xavier Holt

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The problem is that it is most likely much slower, due to traversal of the hirarchy. – Xeo Jun 01 '11 at 18:40
@Xeo - What would make this slow? A smart compiler should be able to optimize out the function calls, leaving you with a mere two pointer dereferences. – Xavier Holt Jun 01 '11 at 18:57
I meant the `dynamic_cast` or `typeid` approach. :) – Xeo Jun 01 '11 at 18:57

score 8 · Answer 2 · edited May 14 '19 at 15:03

8

I would suggest using qobject_cast https://doc.qt.io/qt-5/qobject.html#qobject_cast

It works like dynamic_cast but is a little better since it can make some Qt specific assumptions (doesn't depend on RTTI).

You can use it like this:

if(QPushButton *pb = qobject_cast<QPushButton*>(widget)) {
    // it's a "QPushButton", do something with pb here
}
// etc

edited May 14 '19 at 15:03

Vaidotas Strazdas

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answered Jun 01 '11 at 18:51

Evan Teran

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score 1 · Answer 3 · edited May 14 '19 at 18:16

1

You can write following utility functions:

bool IsCheckBox(const QWidget *widget)
{
   return dynamic_cast<const QCheckBox*>(widget) != 0;
}
bool IsComboBox(const QWidget *widget)
{
   return dynamic_cast<const QComboBox*>(widget) != 0;
}

Or maybe, you can use typeid to determine the runtime type of the object in the cell.

EDIT:

As @Evan noted in the comment, you can also use qobject_cast to cast the object, instead of dynamic_cast. See the examples here.

edited May 14 '19 at 18:16

Vaidotas Strazdas

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answered Jun 01 '11 at 18:24

Nawaz

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better to use `qobject_cast`. It is faster, doesn't depend on RTTI and works across library boundaries. – Evan Teran Jun 01 '11 at 18:49

score 0 · Answer 4 · answered Jun 01 '11 at 18:36

0

You can use QObject::className() to get the type of the widget.

answered Jun 01 '11 at 18:36

BЈовић

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How to determine type of widget in a qtable cell?

4 Answers4