在 AVR 中处理多个 PCINT 的最佳方式
Best way to handle multiple PCINT in AVR
我正在 Attiny85 上测试一些东西,并考虑了处理中断例程的最佳方法。我知道在中断处理程序中有很多代码是不好的,但我不确定是否有其他方法可以做到这一点。我希望我的主程序在 PCINT 上睡眠和唤醒,PCINT 来自多个引脚(旋转编码器 A、b 和开关以及一个接收 UART)所以我想在处理程序中只需要很多代码。
确定哪个引脚引起中断的代码如下所示
#include <avr/io.h>
#include <stdint.h> // has to be added to use uint8_t
#include <avr/interrupt.h> // Needed to use interrupts
volatile uint8_t portbhistory = 0xFF; // default is high because the pull-up
int main(void)
{
DDRB &= ~((1 << DDB0) | (1 << DDB1) | (1 << DDB2)); // Clear the PB0, PB1, PB2 pin
// PB0,PB1,PB2 (PCINT0, PCINT1, PCINT2 pin) are now inputs
PORTB |= ((1 << PORTB0) | (1 << PORTB1) | (1 << PORTB2)); // turn On the Pull-up
// PB0, PB1 and PB2 are now inputs with pull-up enabled
PCICR |= (1 << PCIE0); // set PCIE0 to enable PCMSK0 scan
PCMSK0 |= (1 << PCINT0); // set PCINT0 to trigger an interrupt on state change
sei(); // turn on interrupts
while(1)
{
/*main program loop here */
}
}
ISR (PCINT0_vect)
{
uint8_t changedbits;
changedbits = PINB ^ portbhistory;
portbhistory = PINB;
if(changedbits & (1 << PB0))
{
/* PCINT0 changed */
}
if(changedbits & (1 << PB1))
{
/* PCINT1 changed */
}
if(changedbits & (1 << PB2))
{
/* PCINT2 changed */
}
}
然后在中断处理程序的每个 if 语句中进行 ofc,会有代码处理一些事情,比如这段代码,打开 Timer0
TCNT0 = 0; // Set counter to 0
OCR0A = SERIAL_BIT_TIME; // Call timer interrupt in middle of first bit
position = 0; // Reset position and data
TIMSK |= 1 << OCIE0A; // Enable interrupt for compare register A (timer interrupt)
TIFR |= 1 << OCF0A; // Clear timer interrupt flag to prevent it jumping directly there
PCMSK &= ~(1 << SERIAL_RECEIVE); // Disable pin change interrupt
或者对于开关输入,if 语句中的代码将是
if (lightState)
{
dali.transmit(ADDRESS, OFF);
lightState = 0;
}
else
{
dali.transmit(ADDRESS, ON);
lightState = 1;
}
这是一个愚蠢的解决方案吗?
volatile uint8_t flag;
int main(void)
{
DDRB &= ~((1 << DDB0) | (1 << DDB1) | (1 << DDB2)); // Clear the PB0, PB1, PB2 pin
// PB0,PB1,PB2 (PCINT0, PCINT1, PCINT2 pin) are now inputs
PORTB |= ((1 << PORTB0) | (1 << PORTB1) | (1 << PORTB2)); // turn On the Pull-up
// PB0, PB1 and PB2 are now inputs with pull-up enabled
PCICR |= (1 << PCIE0); // set PCIE0 to enable PCMSK0 scan
PCMSK0 |= (1 << PCINT0); // set PCINT0 to trigger an interrupt on state change
sei(); // turn on interrupts
while(1)
{
gotosleep();
do
{
switch(flag)
{
case 1:
dosomething1();
break;
case 2:
dosomething2();
break;
case 3:
dosomething3();
break;
}
cli();
flag = 0;
sei();
}while(flag);
}
}
ISR (PCINT0_vect)
{
uint8_t changedbits;
changedbits = PINB ^ portbhistory;
portbhistory = PINB;
if(changedbits & (1 << PB0))
{
flag = 1;
}
if(changedbits & (1 << PB1))
{
flag = 2;
}
if(changedbits & (1 << PB2))
{
flag = 3;
}
}
我正在 Attiny85 上测试一些东西,并考虑了处理中断例程的最佳方法。我知道在中断处理程序中有很多代码是不好的,但我不确定是否有其他方法可以做到这一点。我希望我的主程序在 PCINT 上睡眠和唤醒,PCINT 来自多个引脚(旋转编码器 A、b 和开关以及一个接收 UART)所以我想在处理程序中只需要很多代码。
确定哪个引脚引起中断的代码如下所示
#include <avr/io.h>
#include <stdint.h> // has to be added to use uint8_t
#include <avr/interrupt.h> // Needed to use interrupts
volatile uint8_t portbhistory = 0xFF; // default is high because the pull-up
int main(void)
{
DDRB &= ~((1 << DDB0) | (1 << DDB1) | (1 << DDB2)); // Clear the PB0, PB1, PB2 pin
// PB0,PB1,PB2 (PCINT0, PCINT1, PCINT2 pin) are now inputs
PORTB |= ((1 << PORTB0) | (1 << PORTB1) | (1 << PORTB2)); // turn On the Pull-up
// PB0, PB1 and PB2 are now inputs with pull-up enabled
PCICR |= (1 << PCIE0); // set PCIE0 to enable PCMSK0 scan
PCMSK0 |= (1 << PCINT0); // set PCINT0 to trigger an interrupt on state change
sei(); // turn on interrupts
while(1)
{
/*main program loop here */
}
}
ISR (PCINT0_vect)
{
uint8_t changedbits;
changedbits = PINB ^ portbhistory;
portbhistory = PINB;
if(changedbits & (1 << PB0))
{
/* PCINT0 changed */
}
if(changedbits & (1 << PB1))
{
/* PCINT1 changed */
}
if(changedbits & (1 << PB2))
{
/* PCINT2 changed */
}
}
然后在中断处理程序的每个 if 语句中进行 ofc,会有代码处理一些事情,比如这段代码,打开 Timer0
TCNT0 = 0; // Set counter to 0
OCR0A = SERIAL_BIT_TIME; // Call timer interrupt in middle of first bit
position = 0; // Reset position and data
TIMSK |= 1 << OCIE0A; // Enable interrupt for compare register A (timer interrupt)
TIFR |= 1 << OCF0A; // Clear timer interrupt flag to prevent it jumping directly there
PCMSK &= ~(1 << SERIAL_RECEIVE); // Disable pin change interrupt
或者对于开关输入,if 语句中的代码将是
if (lightState)
{
dali.transmit(ADDRESS, OFF);
lightState = 0;
}
else
{
dali.transmit(ADDRESS, ON);
lightState = 1;
}
这是一个愚蠢的解决方案吗?
volatile uint8_t flag;
int main(void)
{
DDRB &= ~((1 << DDB0) | (1 << DDB1) | (1 << DDB2)); // Clear the PB0, PB1, PB2 pin
// PB0,PB1,PB2 (PCINT0, PCINT1, PCINT2 pin) are now inputs
PORTB |= ((1 << PORTB0) | (1 << PORTB1) | (1 << PORTB2)); // turn On the Pull-up
// PB0, PB1 and PB2 are now inputs with pull-up enabled
PCICR |= (1 << PCIE0); // set PCIE0 to enable PCMSK0 scan
PCMSK0 |= (1 << PCINT0); // set PCINT0 to trigger an interrupt on state change
sei(); // turn on interrupts
while(1)
{
gotosleep();
do
{
switch(flag)
{
case 1:
dosomething1();
break;
case 2:
dosomething2();
break;
case 3:
dosomething3();
break;
}
cli();
flag = 0;
sei();
}while(flag);
}
}
ISR (PCINT0_vect)
{
uint8_t changedbits;
changedbits = PINB ^ portbhistory;
portbhistory = PINB;
if(changedbits & (1 << PB0))
{
flag = 1;
}
if(changedbits & (1 << PB1))
{
flag = 2;
}
if(changedbits & (1 << PB2))
{
flag = 3;
}
}