中断是由启用中断之前发生的更改引起的

Question

I'm using an Arduino Uno with the Atmega328p microcontroller. I'm trying to use INT1 as a software interrupt. I manually set INT1's associated PORTD3 high or low depending on external info. What I want is to set the pin high or low at the startup of the device and then enable the interrupt on the pin without causing an interrupt if i set the pin high before enabling the interrupt.

It doesnt seem to matter where I enabled the interrupt--if i changed the state of the pin at some point the interrupt will occur once its enabled. Here is a snippet of the code:

int main(void)
{
    DDRD |= (1<<DDD7)|(1<<DDD3);//7 for siren 3 for software int1
    USART_Init(MYUBRR);//Initialize USART
    while(door!='C'  && door!='O'){//get door state on startup
        door = getDoorState();
    }
    if(door=='O')
        PORTD |= 1<<PORTD3;
    else
        PORTD &= ~(1<<PORTD3);
    EIFR &= ~(1<<INTF1);//clear flag bit before enable, I'd heard this may help????
    EIMSK |= (1<<INT1);//enable door switch interrupt
    EICRA |= (1<<ISC00)|(1<<ISC10);//int1 and int0 set for any logical change

    sei();//global interrupt enable

    while (1) 
        {}
}

As soon as the global interrupt is enabled by a call to sei() the interrupt will occur if PORTD3 is high, regardless of where PORTD3 was set high or where sei() is. Calling sei() should never cause an interrupt in this code, ideally.

Answer 1

4386427 is correct. The bit is cleared by setting it to a one, not zero. Seems counter-intuitive to me so it threw me off but it works now.

EIFR |= (1<<INTF1);

Answer 2

EIFR &= ~(1<<INTF1) is incorrect.

The correct way of doing this is EIFR = 1<<INTF1 .

Datasheet says: the flag is cleared by writing a '1' to it.