Signal is not read by the Class

[Vijaykarthikeyan]

@jsulm Can u guide me to anyother link where I can completely learn that in Qt creator platform..

[Christian Ehrlicher]

@Vijaykarthikeyan said in Signal is not read by the Class:

Qt creator platform

Qt Creator is an IDE, no platform.
Searching by yourself for c++ beginners stuff in the net should not be that hard.

[Vijaykarthikeyan]

@Christian-Ehrlicher But,before this..i followed the same procedure for UDP communication.It worked fine..The data received from udp accessed by the second class..That's why I have tried like this.I have nothing changed a single line from the udp communciation..If the data accessed by second class in udp by this method, why not working in this serial communication,,That's why I have doubted baudrate.

[jsulm]

@Vijaykarthikeyan It is really not difficult to pass a pointer to MainWindow (or any other data) to your Link class:

class Link:
{
public:
    Link(MainWindow *mainWindow_):
        mainWindow(mainWindow_)
    {}

[J.Hilk]

So, @Vijaykarthikeyan for your better understanding:

Here's a simple example to illustrate the difference between a C++ file and an instance of a class.

Let's say we have a C++ file named Dog.cpp that defines a Dog class:

// Dog.cpp
#include <iostream>
#include <string>

class Dog {
public:
    std::string name;

    Dog(std::string dogName) {
        name = dogName;
    }

    void bark() {
        std::cout << name << " says: Woof!" << std::endl;
    }
};

In this file, we've defined a Dog class with a constructor and a bark method. This is just a blueprint for creating Dog objects, not an instance itself.

Now, in another C++ file (let's call it main.cpp), we can create multiple instances of the Dog class:

// main.cpp
#include "Dog.cpp"

int main() {
    Dog dog1("Fido");
    Dog dog2("Rex");

    dog1.bark();  // Outputs: Fido says: Woof!
    dog2.bark();  // Outputs: Rex says: Woof!

    return 0;
}

In main.cpp, we've created two instances of the Dog class, each with its own name. Each instance (dog1 and dog2) is independent of the other. They are both instances of the Dog class, but they are separate objects with their own state (in this case, their names).

So, one C++ file (Dog.cpp) does not equate to one instance of an object/class. The C++ file contains the definition (or blueprint) of the class, and instances of that class can be created in any other part of the program (like in main.cpp).

---

Lets modify the example to include another class that takes a Dog pointer and calls the bark method. Let's create a new class called Person:

// Person.cpp
#include "Dog.cpp"

class Person {
public:
    std::string name;

    Person(std::string personName) {
        name = personName;
    }

    void command(Dog* dog) {
        std::cout << name << " commands: ";
        dog->bark();
    }
};

In this Person class, we have a command method that takes a pointer to a Dog object and calls the Dog's bark method.

Now, let's modify our main.cpp to use this new Person class:

// main.cpp
#include "Person.cpp"

int main() {
    Dog dog1("Fido");
    Person person1("Alice");

    person1.command(&dog1);  // Outputs: Alice commands: Fido says: Woof!

    return 0;
}

In this modified main.cpp, we create a Dog instance (dog1) and a Person instance (person1). We then call the command method on the Person instance, passing in a pointer to the Dog instance. This results in the Person instance commanding the Dog instance to bark. I hope this helps! Let me know if you have any other questions.

[Vijaykarthikeyan]

@J-Hilk ok.. I understand that instead of instantiating the class, we should pass the class member as a pointer to the 2nd class method. So that should be the correct procedure.

[J.Hilk]

@Vijaykarthikeyan said in Signal is not read by the Class:

@J-Hilk ok.. I understand that instead of instantiating the class, we should pass the class member as a pointer to the 2nd class method. So that should be the correct procedure.

I can't quite sign that statement the way it is ....

Yes, if you do it that way, it will work.

What you "should" do is make the QObject::connect in the scope that instantiates your Link class

[Vijaykarthikeyan]

@J-Hilk like this??

Link::Link(MainWindow *mainWindow_): mainWindow(mainWindow_)
{
connect(mainWindow,&MainWindow::data_received,this,&Link::update);
}

[J.Hilk]

@Vijaykarthikeyan

[Vijaykarthikeyan]

@J-Hilk I have another question which i have posted in another link..can you explain that too?

[jsulm]

@Vijaykarthikeyan You could at least post that another link...

[Vijaykarthikeyan]

@jsulm https://forum.qt.io/post/775285