How to get global mouse position in Qt6? - c++

I'm trying to get global position of the mouse however the functions mentioned in older topics are either deprecated or not working as intended. I have tried QCursor::pos() as seen below but it didn't work as intended.
#include <QtCore/QCoreApplication>
#include <QCursor>
#include <iostream>
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
while(1)
{
std::cout << QCursor::pos().x() << " " << QCursor::pos().y() << std::endl;
}
return a.exec();
}
Output: 2147483647 2147483647

It is very simple. QCursor is in the Gui module of Qt see here, so you have to change QCoreApplication to QGuiApplication and then it works (already tested it).

Related

Purpose of QcoreApplication in basic Qt console program

Why does QCoreApplication a(argc, argv) exist inside main(int argc, char *argv[]) in a Qt console application? What does it do? What is its purpose?
if I remove it, a program like the following runs without issues
#include <QCoreApplication>
#include "animal.h"
int main(int argc, char *argv[])
{
//QCoreApplication a(argc, argv);
{ // this is just a dummy scope for the destructor to work
// create instances of class animal
animal cat(NULL, "Garfield");
animal dog(NULL, "Azor");
// perform actions with the object's functions
cat.weight = 10;
qDebug() << "Cat Weight in Lbs:" << cat.weightInLbs();
cat.speak("meow");
// perform actions with the object's functions
dog.weight = 35;
qDebug() << "Dog Weight in Lbs:" << dog.weightInLbs();
dog.speak("wooof");
}
//return a.exec();
return 0;
}

Incomplete Type is not Allowed QApplication

I am learning C++ with Qt. I have installed Qt 5.15 and am using VS2019. I have the below code (as per an example in a textbook I am working through):
#include <QtGui>
int main(int argc, char* argv[])
{
QApplication app(argc, argv);
QTextStream cout(stdout);
//declarations of variables
int answer = 0;
do{
//local variables
int factArg = 0;
int fact(1);
factArg = QInputDialog::getInt(0, "Factorial Calculator", "Factorial of:", 1);
cout << "User entered: " << factArg << endl;
int i = 2;
while (i <= factArg) {
fact = fact * i;
++i;
}
QString response = QString("The factorial of %1 is %2.\n%3").arg(factArg).arg(fact).arg("Do you want to compute another factorial?");
answer = QMessageBox::question(0, "Play again?", response, QMessageBox::Yes | QMessageBox::No);
} while (answer == QMessageBox::Yes);
return EXIT_SUCCESS;
}
However, I am recieving the below error when creating an instance of QApplication as app:
Incomplete Type is not Allowed
I am also recieving the below error for the QInputDialog and QMessageBox classes:
name followed by '::' must be a class or namespace name
I am not sure why this is happening - presumably something with a namespace, but I am not sure what scope to provide. I have searched the web but to no avail.
UPDATE
Adding the below header references give cannot open source file error for each.
#include <QApplication>
#include <QInputDialog>
#include <QMessageBox>
I have also added suggestions from the comments to my code, now below:
#include <QtGui>
#include <iostream>
using namespace std;
int main(int argc, char* argv[])
{
QApplication app(argc, argv);
QTextStream cout(stdout);
//declarations of variables
int answer = 0;
do{
//local variables
int factArg = 0;
int fact(1);
factArg = QInputDialog::getInt(0, "Factorial Calculator", "Factorial of:", 1);
cout << "User entered: " << factArg << endl;
int i = 2;
while (i <= factArg) {
fact = fact * i;
++i;
}
QString response = QString("The factorial of %1 is %2.\n%3").arg(factArg).arg(fact).arg("Do you want to compute another factorial?");
answer = QMessageBox::question(0, "Play again?", response, QMessageBox::Yes | QMessageBox::No);
} while (answer == QMessageBox::Yes);
return EXIT_SUCCESS;
}
But I am still recieving the same errors.
You have to include some qt headers in the app... that is the meaning of the message
you just need to add this to your code
#include <QApplication>
#include <QInputDialog>
#include <QMessageBox>
The correct headers to include are the following:
#include <QtWidgets/QApplication>
#include <QtWidgets/QInputDialog>
#include <QtWidgets/QMessageBox>
Once declaring these, the compiler accepts the code.
I'm going to cut and paste a working program for you.
Qt seems to have a bunch of forward definitions. I can tell I'm missing the right include file if I try to get the IDE to actually help me select a method call. In short, you pretty much have to #include every widget type or other class you're going to use. I haven't run into any shortcuts.
#include <QApplication>
#include "MainWindow.h"
using namespace std;
/**
* Main entry point.
*/
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
MainWindow mainWindow;
mainWindow.show();
return app.exec();
}

Why messages printed after closing main window in Qt C++?

With this code
#include "mainwindow.h"
#include <QApplication>
#include <iostream>
#include <QDir>
#include <QTextStream>
int main(int argc, char *argv[]){
QApplication a(argc, argv);
QTextStream out(stdout);
out << QDir::currentPath();
std::cout << "Why is that?";
MainWindow mainWindow;
mainWindow.show();
return a.exec();
}
Both messages printed only after closing Main Window of my app, why is this?
I tried to debug, debugger thinks that he done with this line, but I see no messages.
extern std::ostream cout; is buffered, so it may choose when to flush its buffer to stdout. In your case, it is doing it when your program is terminating.
You can tell std::ostream to flush using std::flush, as such:
std::cout << "Why is that?" << std::flush;

QApplication changing value of std::to_string

It seems that Qt changes the 'interpretation' of the returned string of std::to_string(double).
Example:
#include <iostream>
#include <QApplication>
#define VERSION 1.0
#define VSIZE 3
int main(int argc, char** argv) {
std::string version = std::to_string(VERSION).substr(0, VSIZE);
std::cout << version << std::endl; // everything fine
QApplication app(argc, argv);
std::cout << std::to_string(VERSION).substr(0, VSIZE); // no good
return 0;
}
The output is:
1.0
1,0
I think it's Qt because this is the thinnest example where this happens, and removing the QApplication declaration makes it work again. How can I fix this?

Using std::cout in Qt Gui

I have an application that when run through terminal, the user has the option between command-line mode or GUI mode.
There doesn't seem to be any output to the console at all when using std::cout. std::cout statements don't work in the main event loop.
I have added CONFIG += console to my .pro file.
For now, I have been using QTextStream() which works fine:
QTextStream(cout) << "Hello World" << std::endl;
My question is:
Why can I not use std::cout? Does this have something to do with Qt affecting input and output streams? I couldn't find any documentation in Qt's docs on this.
int main(int argc, char *argv[])
{
std::cout << argv[1] << std::endl; //This is being outputted.
//if(argc == 2 && !strcmp(argv[1],"-win")){
if(true){ //Just for this example's sake
QApplication a(argc, argv);
std::cout << "Hello" << std::endl; //This is not being ouputted.
MainWindow w;
w.show();
return a.exec();
}
else
{
qDebug() << "Console Mode.\n";
std::cout << "Console Mode.\n";
//Do stuff
}
}
This is not a Qt issue, but how std::cout works. You seem to blow up your std::cout in here:
std::cout << argv[1] << std::endl;
Your issue can be reproduced even with a simple program like this:
main.pro
TEMPLATE = app
TARGET = main
CONFIG -= qt
SOURCES += main.cpp
main.cpp
#include <iostream>
int main(int /*argc*/, char **argv)
{
std::cout << argv[1] << std::endl;
std::cout << "Hello stdout!" << std::endl;
if (std::cout.bad())
std::cerr << "I/O error while reading\n";
return 0;
}
Build and Run
Success: qmake && make && ./main foo
Failure: qmake && make && ./main
In your case argv[1] is nil and so this makes std::cout not to print anything more. I would suggest to either pass an argument all the time and/or check against argc with some help usage print. The best would be to use the builtin command line parser in QtCore these days.
You could ask why? Because it is undefined behavior. You can read the details from the documentation:
basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);
After constructing and checking the sentry object, checks if sb is a null pointer. If it is, executes setstate(badbit) and exits.
If you happen to have an issue with the IDE itself, for instance QtCreator, then follow these steps in case of QtCreator:
Projects -> Select a kit -> Run tab -> Run section -> Arguments
Works OK for me:
QT += core
QT -= gui
TARGET = untitled
CONFIG += console
CONFIG -= app_bundle
TEMPLATE = app
SOURCES += main.cpp
main.cpp:
#include <QCoreApplication>
#include <QTextStream>
#include <iostream>
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
std::cout << "Hello World" << std::endl;
return a.exec();
}
EDIT:
#include <QCoreApplication>
#include <QTextStream>
#include <QtWidgets/QWidget>
#include <QDebug>
#include <iostream>
int main(int argc, char *argv[]) {
std::cout << "test" << std::endl; // <--- THE PROBLEM IS HERE...IF YOU TRY A SIMPLE STRING IT WORKS FINE SO THE PROBLEM IS argv[1] IS AN EMPTY STRING
//if(argc == 2 && !strcmp(argv[1],"-win")){
if(true){
//Just for this example's sake
QCoreApplication a(argc, argv);
std::cout << "Hello" << std::endl; //This is not being ouputted.
return a.exec();
}
else
{
qDebug() << "Console Mode.\n";
std::cout << "Console Mode.\n";
//Do stuff
} }