How to "emit " SIGNAL " after "addItem"?

Anonymous_Banned275

@KroMignon said in How to "emit " SIGNAL " after "addItem"?:

@AnneRanch I cannot really understand what problem you try to solve.

Your explanation are very confusing for me.
You can add as many signals (or slots) you need to your class. And when signal is emitted is up to you.

Just for clarification:

a SIGNAL is only a function, which is created at compilation during MOC process (you can find it implementation in <class_name>_moc.cpp).

emit is just a semantical sugar only to help programmer be aware about calling a SIGNAL funct> - emit is just a semantical sugar only to help programmer be aware about calling a SIGNAL function.ion.

Qt is not "real time ", so what do you mean with this?

The "display " is part of SIGNAL / SLOT and its real time varies .

emit is just a semantical sugar only to help programmer be aware about calling a SIGNAL function.

Not according to this

JonB

@AnneRanch
As we have discussed before you see that the definition of emit in Qt header file is:

#define emit

See https://code.woboq.org/qt5/qtbase/src/corelib/kernel/qobjectdefs.h.html#_M/emit. That's it. So it is "semantical sugar".

So your clazy message shows it looks for whether you place emit before calling a signal function, and just warns you if you do not. The code still works unaffected without, it's just trying to make your life better, possibly.

fcarney

"real time" has a very specific meaning in embedded and controls programming.
It means a system can satisfy deterministic constraints. This means if the constraint is 100mS then the system will always do certain functions every 100mS. Nit-picky I know, but it can cause confusion in technical discussions.

By your usage I think you mean "the time it takes" for the operation to complete.

Anonymous_Banned275

@fcarney Appreciate your clarification.
I guess typical case "you are dammed if you do or do not ".
I tried to give my question some background ( some people need ) and see how it got away from the real issue.

Precise terminology is useful, but the message Is
"missing ... "and it does NOT imply "warning " to me .....

JonB

@AnneRanch said in How to "emit " SIGNAL " after "addItem"?:

"missing ... "and it does NOT imply "warning " to me .....

"Missing" implies "missing". That has nothing to do with whether something is an error, a warning, a informational message, a hint or a philosophical musing. On the other hand, doesn't the yellow colored triangle and text tell you it's a warning, and red color indicates an error? What is the point of you often stating you do not like the wording of messages? They are what they are.

If you care, https://www.kdab.com/nailing-13-signal-slot-mistakes-clazy-1-3/ states:

11. incorrect-emit

For readability purposes you should always use emit (or Q_EMIT) when calling a signal. Conversely, you should not use those macros when calling something other than a signal.

Clazy will warn if you forget to use emit (or Q_EMIT) or if you use them when you shouldn’t.

which I would have thought is the actually important information.

Anonymous_Banned275

OK, I am still not getting it.
My syntax is obviously wrong.

   emit pass_item();        works 
   emit mySignal(item); does not work

// signals:
// void mySignal(QListWidgetItem * item);

    connect(this,
            SIGNAL(mySignal(item)),
            this ,
            SLOT(testSlot()));

This works fine

    connect(this,
            SIGNAL(pass_item()),
            this ,
            SLOT(testSlot_PASS_DATA() ));

And here is my debug / run time error

TASK : test local SIGNAL / SLOT connect

File : ../../../JUNE 9 WORKCOPY 1/CAT/btscanner/device.cpp
Function : testSlot
@line : 380
count : 1
QObject::connect: No such signal DeviceDiscoveryDialog::mySignal(item) in ../../../JUNE 9 WORKCOPY 1/CAT/btscanner/device.cpp:148
QObject::connect: (sender name: 'DeviceDiscovery')
QObject::connect: (receiver name: 'DeviceDiscovery')

TASK : test local SIGNAL / SLOT connect

File : ../../../JUNE 9 WORKCOPY 1/CAT/btscanner/device.cpp
Function : testSlot
@line : 380
count : 2

JonB

@AnneRanch said in How to "emit " SIGNAL " after "addItem"?:

SIGNAL(mySignal(item)),

When connecting signals this (old) SIGNAL macro way, you need to specify the type(s) of the parameter(s):

SIGNAL(mySignal(QListWidgetItem *))

KroMignon

@AnneRanch You should avoid using old connect syntax and use new one:

 connect(this,
            SIGNAL(mySignal(item)),
            this ,
            SLOT(testSlot()));

becommes:

 connect(this,
            &ClassName::mySignal,
            this ,
            &ClassName::testSlot);

So you will avoid this kind of errors.

JonB

@KroMignon
We have suggested this many, many times.....! :)

VRonin

It's the exact same problem you reported here. You can apply the same solution

Anonymous_Banned275

@VRonin said in How to "emit " SIGNAL " after "addItem"?:

It's the exact same problem you reported here. You can apply the same solution

Yes, same issue.
I think if somebody said "pointer to function" it may have help.

Now as far as "passing the data" the syntax looks as standard passing parameter to a function passed as a parameter.
Sounds silly but that is what it is , nothing Qt special when implementing "connect".

Anonymous_Banned275

@JonB said in How to "emit " SIGNAL " after "addItem"?:

@KroMignon
We have suggested this many, many times.....! :)
Thanks for keeping track
and you may expect more of the same in the future - so ignore it.

Anonymous_Banned275

OK, few more questions , if I am allowed to push my luck here.

The SIGNAL function has parameter "item" , then
the SLOT function should have same type of parameter?

BUT the new connect syntax does not require the parameters, hence in "connect" neither SIGNAL or SLOT functions are with parameters, and it passes MOC fine.

How do I process the undeclared parameter in my SLOT function?

BUT If I declare the SLOT with parameter my code breaks.

Again if the above questions are bothersome / repetitious just ignore my post.

  emit pass_item();
   emit mySignal(item);

// signals:
// void mySignal(QListWidgetItem * item);
// emit testSlot_PASS_DATA ();

    connect(this,
                &DeviceDiscoveryDialog::mySignal,
                this ,
                &DeviceDiscoveryDialog::testSlot);

    connect(this,
            &DeviceDiscoveryDialog::mySignal,
            this ,
            &DeviceDiscoveryDialog::testSlot_PASS_DATA);

artwaw

@AnneRanch if your slot doesn't require parameter it's ok. If you need to catch the parameter but can't provide/modify corresponding slot for use with parameter, you can still catch it using lambda (using new connect() syntax. And then do something with that parameter, call helper method or whatever and slot at the beginning/end, however you like.

Anonymous_Banned275

Now that is clear as mud.
I have control over the SLOT - that is not the issue.
My question is
if the SIGNAL contains , carries or whatever term is used , the "item" (as a parameter) without specifying "item' in ":connect" as parameter
AND
SLOT connects to such SIGNAL , again without specifying it in "connect"
syntax
HOW DO I ACCESS THE "item" passed (?) by SIGNAL in my SLOT function ?
So far all I have done is to activate the SLOT function - now I need to learn how to use the parameter carried in by SIGNAL.

SGaist

Hi,

Did you go through the Signals and Slots chapter in Qt's documentation ?
It explains everything and provides simple examples to show how things work.

Anonymous_Banned275

@SGaist Just scanned thru thru this .

https://wiki.qt.io/New_Signal_Slot_Syntax

Did not find a single example how SIGNAL with parameter is used in SLOT.
It does not explain how such parameter gets thru to the SLOT also.
There is no SLOT function example either.
I'll keep looking , there got to be a pony somewhere...

SGaist

The Counter class from the documentation:

#include <QObject>

class Counter : public QObject
{
    Q_OBJECT

public:
    Counter() { m_value = 0; }

    int value() const { return m_value; }

public slots:
    void setValue(int value);

signals:
    void valueChanged(int newValue);

private:
    int m_value;
};

The setValue slot:

void Counter::setValue(int value)
{
    if (value != m_value) {
        m_value = value;
        emit valueChanged(value);
    }
}

The connection:

 Counter a, b;
    QObject::connect(&a, &Counter::valueChanged,
                     &b, &Counter::setValue);

The usage:

    a.setValue(12);     // a.value() == 12, b.value() == 12
    b.setValue(48);     // a.value() == 12, b.value() == 4

When you call setValue from object a and that this value does not match the current counter, the valueChanged signal will be emitted with the new value as parameter.
Then, the setValue of b will be called with as parameter the new value.

ChrisW67

@AnneRanch said in How to "emit " SIGNAL " after "addItem"?:

It does not explain how such parameter gets thru to the SLOT also.

When you are using the new style signal-slot connections then the C++ compiler checks that the sender and receiver have matching signatures (or close enough) at compile time and fails to compile if they do not.
https://wiki.qt.io/New_Signal_Slot_Syntax#Type_mismatch

There's not much to explain here to the user of the signal-slot mechanism. The value passed with the signal function call, e.g. fooSignal("barbaz"), is delivered to any slot function connected to the signal essentially as if you had called fooSlot("barbaz"). (The slot may not be called instantly depending on circumstances.) The precise internals of this mechanism are much more involved, but only of relevance if you intend to develop the Qt library yourself.

KroMignon

@AnneRanch said in How to "emit " SIGNAL " after "addItem"?:

Now that is clear as mud.
I have control over the SLOT - that is not the issue.

My question is
if the SIGNAL contains , carries or whatever term is used , the "item" (as a parameter) without specifying "item' in ":connect" as parameter
AND
SLOT connects to such SIGNAL , again without specifying it in "connect"
syntax

This don't made sense to me.

Qt has 2 syntax to create a connection:

the "old" one with SLOT() and SIGNAL() macros
the "new" one where you can specify the signal and slots functions.

With the old syntax, you have to specify all parameter types (not names), because corresponding slot or signal are found per introspection (all QMeta stuff). So there is no way, at compilation time, to know if signal or slot exists.

With the new syntax, you give full function name. This way it is possible, at compilation, to the check if corresponding signal/slots exists.
With the new syntax you can also connection a signal to a functor/lambda function. Which is not possible with the old syntax.

HOW DO I ACCESS THE "item" passed (?) by SIGNAL in my SLOT function ?
So far all I have done is to activate the SLOT function - now I need to learn how to use the parameter carried in by SIGNAL.

For all of them, signal and slot do not have to have the same amount of parameters, but signal must at least have same type of parameters as connected slot.

Suppose you have follow signals:

void statusChanged(int newStatus)
void newMessage(const QString &message)
void done()

And following slots:

void onStatusChanged(int newStatus)
void onStatusMessageChanged(int newStatus, const QString &message)
void writeToLog(const QString &message)
void triggerTimer()

with old syntax connection would be:

connect(this, SIGNAL(statusChanged(int)), this, SLOT(onStatusChanged(int)));
connect(this, SIGNAL(statusChanged(int)), this, SLOT(triggerTimer()));

same with new syntax:

connect(this, &MyClass::statusChanged, this, &MyClass::onStatusChanged);
connect(this, &MyClass::statusChanged, this, &MyClass::triggerTimer);

But connecting statusChanged with writeToLog will not be possible or onStatusMessageChanged, because parameter types are not compatible.

It seems you are always asking same kind of questions about signals/slots.
There are so many documentation about this

Signals & Slots at doc.qt.io
What do I do if a slot is not invoked? at KDAB
How Qt Signals and Slots Work at woboq

I know, reading documentation is boring, but always having same issue is even more boring.
Don't you think so?