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long long in interrupts: how to write safe in main? (disable int or semaphore?)

Question asked by bil.til on Nov 24, 2012
Latest reply on Nov 24, 2012 by bil.til
Hi,
please have a look at the following code (STM32F407):
volatile long long vllPosition;

void TIM1_UP_TIM10_IRQHandler(void){
... do interrupt stuff ...
vllPosition ++;
... do interrupt stuff ...
}

int main( void){

  ... do some init stuff ...

  for(;;){
    ... do the loop stuff ...

    if( ...){        // from time to time it is necessary to assing some value to llPosition
      vllPosition= llSomeValue;
    }

    ... do the loop stuff ...
  }
}

As long long is 64 bit, the access to vllPosition is non-atomic. So an assignment like vllPosition= llSomeValue is dangerous: it might be interrupted by my interrupt and then the complete thing will mix up.

In an easy access, I would like to disable the interrupt temporarily. There would be 3 different possibilities I see:
1.:
      NVIC_DisableIRQ( TIM1_UP_TIM10_IRQn);
      vllPosition= llSomeValue;
      NVIC_EnableIRQ( TIM1_UP_TIM10_IRQn);

2. (The TIM1_UP_TIM10_IRQ is entered exclusively by the UIE):
      TIM1->DIER = 0;
      vllPosition= llSomeValue;
      TIM1->DIER = TIM_DIER_UIE;
 
3.
      __disable_irq();
      vllPosition= llSomeValue;
      __enable_irq();
 
Is any of these methods 100% safe (of course I would prefer method 1 or 2, as 3 disables all interrupts)? (Due to the prefetch delay I am not sure about this - I did not find a guarantee anywhere and it is difficult to decide only by testing). (Or would I need some more elaborate methode to ensure this in a secure way, e. g. with semaphores or so?)

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