问题描述
我根据新的连接语法在Qt5中使用了以下语法,以避免插槽和信号的类型不匹配对于带有可检查项的 QListWidget
.
I used below syntax in Qt5 according to new connect syntax to avoid type mismatches of slot and signals for a a QListWidget
with checkable items.
connect(item, &QListWidget::itemChanged,this , &mainWindow::checkItemChanged);
我想运行我的插槽,以防任何列表项更改其状态.为此,由于 this answer,我使用了 itemChanged
信号,但它受到保护并且编译时错误如下:
I want to run my slot in case any of list item changed its state. In order to this this I used itemChanged
signal due to this answer, but it is protected and compile time error raise as below:
error: ‘void QListWidget::itemChanged(QListWidgetItem*)’ is protected
我该如何处理?我应该继承我自己的 QListWidget
还是有其他一些解决方案?
How can I handles this? Should I subclass my own QListWidget
or there are some other solutions to this?
推荐答案
可以根据Qt版本使用更合适的语法:
You can use the more appropriate syntax according to Qt version:
#if QT_VERSION >= 0x050000
connect(item, &QListWidget::itemChanged, this , &MainWindow::checkItemChanged);
#else
connect(item, SIGNAL(checkItemChanged), this , SLOT(checkItemChanged));
#endif
(或所有版本的旧的基于字符串").
(or the 'old string-based' for all versions).
这篇关于Qt:将受保护的 QListWidget::itemChanged 信号连接到插槽的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!