Check if widget is pushbutton

Christian Ehrlicher

@hskoglund no, use proper c++ instead string comparison please.

ODБOï

hskoglund

Yes it's better than string comparison, but it's 2 lines :-)

auto pushButton = qobject_cast<QPushButton*>(widget);
if (nullptr != pushButton)

ODБOï

@hskoglund can't you do?

if(auto pushButton = qobject_cast<QPushButton*>(pushButton ))

hskoglund

@LeLev sure but doesn't the compiler complain about '=' inside an if statement?

ODБOï

@hskoglund i just tested with qt5.14/minGw, it compiles and i don't see any warnings

hskoglund

Ok I guess that eliminates all arguments for the string comparison...

Please_Help_me_D

@hskoglund said in Check if widget is pushbutton:

if (QString("QPushButton") == widget->metaObject()->className())
...

this works, thank you!

But when I try to execute:

auto plot = qobject_cast<QwtPlot*>(widget);

I get an error:
mainwindow.obj:-1: error: LNK2001: unresolved external symbol "public: static struct QMetaObject const QwtPlot::staticMetaObject" (?staticMetaObject@QwtPlot@@2UQMetaObject@@B)

QwtPlot is from QWT library. It is declared as:
class QWT_EXPORT QwtPlot: public QFrame, public QwtPlotDict

Christian Ehrlicher

@Please_Help_me_D Then you forgot to link against the qwt library.

Please_Help_me_D

@Christian-Ehrlicher thank you! I added in .pro file:

DEFINES += QWT_DLL

Could you please explain what this line means? I've found it in internet but I don't understand how it works

Christian Ehrlicher

Even it works you should follow the official documentation: https://qwt.sourceforge.io/qwtinstall.html#COMPILEANDLINKAPP

SimonSchroeder

@LeLev said in Check if widget is pushbutton:

@hskoglund can't you do?

if(auto pushButton = qobject_cast<QPushButton*>(pushButton ))

It looks like several C++ features got mixed up in this post. The original question wasn't too far off:

if (static_cast<QPushButton*>(widget))

The only problem here is that the wrong type of cast was used. static_cast<>() means: "I know what I am doing.", i.e. assume that the cast is always allowed. The correct C++ feature is dynamic_cast<>(). So, a C++-only solution would look like this:

if(dynamic_cast<QPushButton*>(widget))

In the context of Qt it was correctly pointed out that qobject_cast<>() is preferrable over dynamic_cast<>() (as it might be faster and does not need RTTI). A normal replacement of the cast is the easiest solution here:

if(qobject_cast<QPushButton*>(widget))

I am not entirely sure what C++ feature would enable the variant suggested by @LeLev :

if(auto pushButton = qobject_cast<QPushButton*>(pushButton ))

As far as I understand variable declarations are not allowed in if statements. C++17 introduced such a feature, however it has the declaration and the condition separated:

if(auto pushButton = qobject_cast<QPushButton*>(pushButton ); pushButton)  // this works since C++17

Has anyone any idea why the version suggested by @LeLev would work?

Bonnie

@SimonSchroeder said in Check if widget is pushbutton:

Has anyone any idea why the version suggested by @LeLev would work?

I've been using this feature for many years, it's definitely NOT a C++17 feature.
As I look into https://en.cppreference.com/w/cpp/language/if , it is called

declaration of a single non-array variable with a brace-or-equals initializer