Timeout

That Demo code is not working on my mbed board. The Led 2 is not flashing every 2 seconds as it should do.

my fault, confusion about timeout and ticker.

Is there an example of using the overloaded method with the: template<typename T> void attach(T * tptr, void (T::*mptr)(void),float t). I am trying to call Timeout inside of a class method but can't figure out the syntax. Here is what I am trying to do: _timeout.attach_us(myClass, myClass::&attimeout, delay); Timeout _timeout is property of the class and attimeout is the method of the same class

Hi Istvan,

Assuming you are trying to attach this in a method within a class, to the instance of itself, try:

_timeout.attach_us(this, &myClass::attimeout, delay);

The format is basically (address of object, address of function within object, delay);

thanks, works perfect

I'm posting this to warn anyone from making the same silly mistake I did. Be careful to check the scope on your Timeout object. If you declare your Timeout object within another function it's possible that the function will resolve and the timeout object will be deleted before your timeout completes.

For instance in this case your blink function will never be called.

if (1){    
        Timeout myTimeout;           //we should declare this object outside of the if statement.
        myTimeout.attach(&blink,2);  //The function blink would be called in two seconds but myTimeout is deleted before then

    }

Hello;

1. include "mbed.h"

Timeout flipper; DigitalOut led1(LED1); DigitalOut led2(LED2);

void flip() { led2 = !led2; }

int main() { led2 = 1; flipper.attach(&flip, 2.0); setup flipper to call flip after 2 seconds

spin in a main loop. flipper will interrupt it to call flip while(1) { led1 = !led1; wait(0.2); } }

I compiled this code but it didn't work for me. I want to make interruptions to execute functions.

THANKS.

Hello; I had a confusion between Ticker and timeout. Now Every thing works fine for me.

:-)

This page should note that the mbed compiler optimises global variables aggressively. So if you use a global variable to communicate between base code and Timeout interrupt it must be marked volatile:

volatile int running = 1;

led1 = DigitalOut(LED1);

void isr_code(void)
{
    running = 0;
}

int main(void)
{
   Timeout.attach(&isr_code, 10.0);
   led1 = 1
   while (running);
   led1 = 0;
   while (1) {
      wait(1);
      led1 = 1 - led1;
   }
}

in this example if running is not volatile the first loop will run forever!

Hello,

a short question:

Is it Ok to call attach() a second/third... time before the timeout has fired or must I call detach() before and attach again?

Michael