我有一个问题,如何在Qt中覆盖信号?
我已经重新定义了QCheckBox()
并更改了stateChanged
信号。
该项目正在运行中。它不会在“应用程序输出”中输出错误,也不会显示“未连接插槽信号”的消息
但是它没有链接到插槽。我不知道怎么了。
此代码有效:
connect(test_checkbox[i], SIGNAL(stateChanged(int)), two_cl , SLOT(run_NEW()));
除了数字之外,我还需要发出字符串:
connect(test_checkbox[i], SIGNAL(stateChanged(int, QString)), two_cl , SLOT(run_NEW(int, QString)));
覆盖QCheckBox
.h
#ifndef MYDIMASCHECKBOX_H
#define MYDIMASCHECKBOX_H
#include <QCheckBox>
class MyDimasCheckBox : public QCheckBox
{
Q_OBJECT
public:
MyDimasCheckBox(QWidget *parent =0);
~MyDimasCheckBox();
QString stroka;
signals:
void stateChanged(int, QString);
};
#endif // MYDIMASCHECKBOX_H
.cpp
#include "mydimascheckbox.h"
MyDimasCheckBox::MyDimasCheckBox(QWidget *parent)
{
stroka = "dimasik :3";
emit stateChanged(int(), stroka);
}
MyDimasCheckBox::~MyDimasCheckBox()
{
}
这是挑战所在
.h
#ifndef MAINWINDOW_H
#define MAINWINDOW_H
#include <QMainWindow>
#include <QCheckBox>
#include <QHBoxLayout>
#include <QDebug>
#include <QThread>
#include <QCoreApplication>
#include <iostream>
#include <vector>
#include "mydimascheckbox.h"
namespace Ui {
class MainWindow;
}
class MainWindow : public QMainWindow
{
Q_OBJECT
int glob_i ;
int glob_flow;
int vector_schet;
public:
explicit MainWindow(QWidget *parent = 0);
~MainWindow();
public slots:
void start_sending(bool);
private slots:
void on_pushButton_clicked();
private:
Ui::MainWindow *ui;
QThread *thread = new QThread();
QVector<QThread*> vector_thread;
QList<MyDimasCheckBox*> test_checkbox;
MyDimasCheckBox *checkBox = new MyDimasCheckBox();
QWidget *checkBoxWidget = new QWidget();
QHBoxLayout *layoutCheckBox = new QHBoxLayout(checkBoxWidget);
};
class NewsThread: public QThread
{
Q_OBJECT
public slots:
void run_NEW();
void run_NEW(int, QString);
signals:
void otprawka (int);
};
#endif // MAINWINDOW_H
.cpp
#include "mainwindow.h"
#include "ui_mainwindow.h"
MainWindow::MainWindow(QWidget *parent) :
QMainWindow(parent),
ui(new Ui::MainWindow)
{
glob_i=0;
glob_flow =0;
vector_schet =0;
ui->setupUi(this);
}
MainWindow::~MainWindow()
{
delete ui;
test_checkbox.clear();
}
void MainWindow::on_pushButton_clicked()
{
glob_i++;
checkBoxWidget = new QWidget();
checkBox = new MyDimasCheckBox();
layoutCheckBox = new QHBoxLayout(checkBoxWidget);
test_checkbox.append(checkBox);
connect(checkBox, SIGNAL(toggled(bool)),this, SLOT(start_sending(bool)));
checkBox->setText(QString::number(glob_i));
layoutCheckBox->addWidget(checkBox);
layoutCheckBox->setAlignment(Qt::AlignCenter);
layoutCheckBox->setContentsMargins(0,0,0,0);
ui->tW_test->insertRow(ui->tW_test->rowCount());
ui->tW_test->setCellWidget(ui->tW_test->rowCount()-1, 1, checkBoxWidget);
qDebug() << "glob_i: " << glob_i;
}
void MainWindow::start_sending(bool Value)
{
qDebug() << "start_sending " ;
// когда нажата отрабатывает, отжата то не отрабатывает
if (Value == true)
{
NewsThread *two_cl = new NewsThread();
qDebug() << "chekbocks: "<< " TRUE" ;
for (int i =0;i < test_checkbox.length();i++ )
{
if(test_checkbox[i]->isChecked() ==Value)
{
glob_flow++;
// connect(test_checkbox[i], SIGNAL(stateChanged(int)), two_cl , SLOT(run_NEW()));
connect(test_checkbox[i], SIGNAL(stateChanged(int, QString)), two_cl , SLOT(run_NEW(int, QString)));
thread = new QThread();
vector_thread.append(thread);
vector_schet++;
qDebug() << "vector_schet : " << vector_schet ;
two_cl->moveToThread(vector_thread[vector_schet-1]);
vector_thread[vector_schet-1]->start();
}
}
}
else {
qDebug() << "chekbocks:" << " False";
glob_flow--;
qDebug() << "vector_schet : " << vector_schet ;
vector_thread[vector_schet-1]->exit();
}
}
void NewsThread::run_NEW()
{
qDebug() << "run_NEW()";
for(;;){
for (int i=0; i<500; i++){
qDebug()<< "Число :" << i <<"number \"flow\" :" ;
usleep(100000);
}
}
}
void NewsThread::run_NEW(int i, QString str){
qDebug() << "run_NEW(int i, QString str) ";
for(;;){
for (int i=0; i<500; i++){
qDebug() << " i : " << i;
qDebug() << " str : " << str;
qDebug()<< "Число :" << i <<"number \"flow\" :" ;
usleep(100000);
}
}
}
答案 0 :(得分:4)
您不能用子类中的另一个信号替换一个信号。但是,您可以在自连接插槽中发出带有原始信号的附加信号:
class MyDimasCheckBox : public QCheckBox
{
Q_OBJECT
public:
MyDimasCheckBox(QWidget *parent =0);
~MyDimasCheckBox();
QString stroka;
private slots:
// Emits the new signal
void doEmitStateChanged(int i);
signals:
void stateChanged(int, QString);
};
MyDimasCheckBox::MyDimasCheckBox(QWidget *parent) : QCheckBox(parent) {
// Connect original signal to slot
connect(this, SIGNAL(stateChanged(int)), this, SLOT(doEmitStateChanged(int)));
}
void MyDimasCheckBox::doEmitStateChanged(int i) {
emit stateChanged(i, stroka);
}
使用新的连接语法,您可以省略插槽并使用lambda:
connect(this, qOverload<int>(&QCheckBox::stateChanged),
// "this" context-object for QThread-Affinity
this, [=](int i) { emit this->stateChanged(i, this->stroka); });
答案 1 :(得分:0)
覆盖信号是一种不良做法[1]:
“具有被覆盖信号的API难以使用,出乎意料且容易出错。更糟糕的是,Qt甚至允许您使用非信号覆盖信号,反之亦然。”
https://www.kdab.com/nailing-13-signal-slot-mistakes-clazy-1-3/