Hi Deoyani,
Next time you may want to use the "code-snippet" </> to show your code :grinning_face:
I have typed what I think you are doing into Notepad++. So here is what I think you are doing in main.
That looks okay to me although I think that the while loop will not do much, temp will perhaps be incremented once or twice. It looks like it was part of a timeout.
disableInterrupts(); //temporarily disable interrupts
GPIO_Init(GPIOE, GPIO_Pin_6, GPIO_Mode_In_FL_IT); //PE6 input, floating, interrupt
EXTI_SetPinSensitivity(EXTI_Pin_6, EXTI_Trigger_Rising_Falling); //EXTI6 on any edge
while (EXTI_GetPinSensitivity(EXTI_Pin_6) != EXTI_Trigger_Rising_Falling)
{ //you just set EXTI_Pin_6 to EXTI_Trigger_Rising_Falling
temp++; //so this will not do much
}
enableInterrupts(); //re-enable interrupts
And here is how I think your interrupt routine is made.
It seems that you clear the interrupt status and after that in the while loop you are checking if it is set!?
It probably will not be, at least not for very long.
@far @interrupt void EXTI6_IRQHandler(void) //unknown (for me) way of handling interrupts
{
uint16_t temp = 0; //not static so at every interrupt set to 0
EXTI_ClearITPendingBit(EXTI_IT_Pin6); //okay, clear interrupt status
while (EXTI_GetITStatus(EXTI_IT_Pin6) != RESET) //but, but, but, you just cleared it!
{
temp++;
}
}
What happens when you make the interrupt routine lke this.
And in main do something when exti_signal == 1 (don't forget to set it to 0 again)
@far @interrupt void EXTI6_IRQHandler(void)
{
if (EXTI_GetITStatus(EXTI_IT_Pin6) != RESET)
{
EXTI_ClearITPendingBit(EXTI_IT_Pin6);
exti_signal = 1;
}
}
Something else... what have you done with all unused GPIO? Don't leave them floating, set them to input with a pull up. If I do not do that, often weird things happen.
