简体   繁体   中英

Cannot connect QButtonGroup buttonClicked to a functor

I am having trouble with connecting QButtonGroup's signal to a functor.

functionGroup = new QButtonGroup();
functionGroup->addButton( ui->pushButtonSetupExperiment, 0);
functionGroup->addButton( ui->pushButtonConfigure, 1);
functionGroup->addButton( ui->pushButtonModify, 2);
functionGroup->addButton( ui->pushButtonSearch, 3);
functionGroup->addButton( ui->pushButtonLogout, 4);

/* This works:
connect( ui->pushButtonSetupExperiment, &QPushButton::clicked, [=]() {
    emit finished(0);
});
*/

// This fails:
connect( functionGroup, &QButtonGroup::buttonClicked, [=](int id) {
    emit finished(id);
});

Following Compiler error occur:

error: C2664: 'QMetaObject::Connection QObject::connect(const QObject *,const char *,const char *,Qt::ConnectionType) const' : Cannot convert argument 2 from 'overloaded-function' to 'const char *'...

Can't figure out what's wrong. Sure I can use the old syntax to complete the task, but I need to learn what's wrong here. Thanks for helping me!

Qt version: 5.5.1
Compiler: msvc2012

The issue is that there are two overloads for QButtonGroup::buttonClicked , and it cannot disambiguate which one you mean, so it generates the compile error. To get around this, you need to use static_cast to indicate which variant of buttonClicked you want to use. Unfortunately, the syntax is a bit clunky:

connect(functionGroup, static_cast<void(QButtonGroup::*)(int)>(&QButtonGroup::buttonClicked), ...);

For more information, see to Differences between Slot Connections docs page as linked by @Devopia. Depending on your personal opinion you might want to use the old string-based connection syntax for this (ie value conciseness and readability over type safety). Or you could use a macro or using type alias to try and simplify things.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM