Call a delegate from another class on interrupt

Projects Programming
1

I have a class that I set to listen to interrupts, and when I get interrupts the method is being called right.
When this class gets interrupt I want to call a method on the main .Ino file so ,

class that gets the interrupt and try to call main Ino method :

Sleep slp;

//set the pointer to the main Ino file
void Sleep::setDelegate( float(*fp)(float,float,char*))
{
    fpAction=fp;
    
}



void Sleep::sleepSystem()
{
...
}

void   wakeOnInt()
{
    
     detachInterrupt(digitalPinToInterrupt(interrupt_pin));
    slp.callMain("INTCOM");
 
 }

void Sleep::callMain(char *arg)
{
 
    char *argt ="INT";
   (*fpAction)(0,0,argt);  //********works till this line - that cause a crash  !
 
}

Then on the main Ino file I have :

//interrupt method declaration
 float interruptDelegate(float varf,float vari,char *arg);

Sleep sleep;

void setup() 
{
        //set the delegate in the sleep class (works)
         sleep.setDelegate(interruptDelegate);
}


//this should be called on interrupt, but it crashes before it
 float interruptDelegate(float varf,float vari,char *arg)
 {
               //Serialr.println(" int  ");
 
 }

As you can see, the interrupt occurs, but when I try to call the main Ino with that method pointer, i get a crash.
Whats wrong with it ?

2

As you can see, the interrupt occurs

Actually, I can't. I see no proof that the interrupt happens.

   (*fpAction)(0,0,argt);  //********works till this line - that cause a crash  !

You know this how?

Usually, before a pointer is dereferenced, one checks that the pointer is valid.

3

"I see no proof that the interrupt happens."

The proof is me, I am telling you this and you have to trust me, otherwise you are just an arrogant guy (which you obviously are)

"You know this how?"

Because I print. Because I remove this line and it works.

Man, you got to trust people, and stop playing this arrogant game.

Its not cool at all.

5

I am not going to absorb this behavior , if you are here to help (and not showing off ) you have to be nice.

I am doing this for a while, and if I say I know its this line, I probably have more information then you to know that because I am on it for a day.

If I don't know , so probably you don't know, and then there is no real logic in this question.

If I remove a line and I print, and I debug, I can tell that line is not good.

Stop being aggressive, I don't care if you are going to help me or not.

6

you have to be nice.

Says who?

I probably have more information then you to know that

I'm sure you do. I hinted, a bit too subtly I see, that you needed to share what only you can see.

Stop being aggressive, I don't care if you are going to help me or not.

Well, OK, then. I'm done. If you don't care if you get help, why did you come here?