[SOLVED] Can't pass object into function

Issue

So I'm trying to make a program to provide a interface between a raspberry pi and a accelerometer/gyroscope. I need this function to update as often as possible and I want to set a timer for when it will run.

And here is my code:

void updateGyro(Gyro gyro) {
    gyro.update();
}

int main() {
    if (gpioInitialise() < 0)
        return -1;

    // Creates new gyro object
    Gyro gyro;
    // Sets a timer to update every 10 milliseconds
    gpioSetTimerFuncEx(0, 10, updateGyro, &gyro);

    

    time_sleep(10);
    gpioTerminate();
}

But this doesn't seem to work and I get the error:

gyro.cpp: In function ‘int main()’:
gyro.cpp:113:31: error: invalid conversion from ‘void (*)(Gyro)’ to ‘gpioTimerFuncEx_t’ {aka ‘void (*)(void*)’} [-fpermissive]
  113 |     gpioSetTimerFuncEx(0, 10, updateGyro, &gyro);
      |                               ^~~~~~~~~~
      |                               |
      |                               void (*)(Gyro)
In file included from gyro.cpp:2:
/usr/include/pigpio.h:3751:55: note:   initializing argument 3 of ‘int gpioSetTimerFuncEx(unsigned int, unsigned int, gpioTimerFuncEx_t, void*)’
 3751 |    unsigned timer, unsigned millis, gpioTimerFuncEx_t f, void *userdata);
      |                                     ~~~~~~~~~~~~~~~~~~^

I've tried passing in the gyro.update function in directly into the function but that also didn't work. Does anyone know how to fix this?

Here is the documentation for the function I want to use:

Solution

void* is not a placeholder for any number of any arguments. It means exactly that, single pointer to unknown type. updateGyro doesn't accept argument of type void*, it accepts argument of type Gyro.

You need to change your function to accept void* instead:

void updateGyro(void* arg) {
    Gyro* gyro = static_cast<Gyro*>(arg);
    gyro->update();
}

Answered By - Yksisarvinen

Answer Checked By - Senaida (WPSolving Volunteer)