Arduino中断对TimeAlarms.h的干扰

Question

我有和arduino的草图，需要使用TimeAlarms.h库在一个定时的时间表上执行几个操作。然而，其中一个操作，通过中断读取霍尔传感器，似乎与TimeAlarms库交互很差。我在这里使用TimeAlarms库：TimeAlarms.html，并从这里修改了霍尔传感器脚本：sensor

我可以自己运行大厅感应器代码。但是，当我试图与Alarm.timerRepeat一起运行霍尔传感器代码时，它会在输入check_flow函数后挂起。

运行下面的代码只输出enter CF，然后挂起。如果您尝试使用check_flow_alarm_delay函数(它使用延迟的TimeAlarm版本)，也会发生同样的情况。

但是，如果您在安装程序中注释掉Alarm.timerRepeat(10, showseconds);，循环中的Alarm.delay(0);则可以很好地工作。

奇怪的是，如果您在sei();函数中注释掉了check_flow和cli();，那么脚本工作得很好，并且似乎在霍尔传感器上进行了正确的计数。为什么要这么做？我是否应该担心，我没有积极地设置sei()和cli()之间的时间，从而导致传感器的可靠性问题？

注意:您应该能够在没有霍尔传感器的情况下运行代码，输出仅为0升/小时。

// reading liquid flow rate using Seeeduino and Water Flow Sensor from Seeedstudio.com
// Code adapted by Charles Gantt from PC Fan RPM code written by Crenn @thebestcasescenario.com
// http:/themakersworkbench.com http://thebestcasescenario.com http://seeedstudio.com

#include <Time.h>
#include <TimeAlarms.h>
#include <Wire.h>


volatile int NbTopsFan; //measuring the rising edges of the signal
int Calc;                               
int hallsensor = 2;    //The pin location of the sensor

void rpm ()     //This is the function that the interupt calls 
{ 
  NbTopsFan++;  //This function measures the rising and falling edge of the hall effect sensors signal
} 

void setup() //
{   
  Serial.begin(9600); //This is the setup function where the serial port is initialised,

  pinMode(hallsensor, INPUT); //initializes digital pin 2 as an input
  attachInterrupt(0, rpm, RISING); //and the interrupt is attached

  Alarm.timerRepeat(10, showseconds);

} 

void loop ()    
{
//  Serial.println( second() );

// stalls at enter CF
//  check_flow();

// stalls at enter CF
  check_flow_alarm_delay();

  Alarm.delay(0);
}

void showseconds ()
{
  Serial.println( second() );
}

void check_flow ()
{
  Serial.println("enter CF");
  int Calc;    
  NbTopsFan = 0;   //Set NbTops to 0 ready for calculations
//  sei();      //Enables interrupts
  delay(1000);   //Wait 1 second
//  cli();      //Disable interrupts
  Calc = (NbTopsFan * 60 / 5.5); //(Pulse frequency x 60) / 5.5Q, = flow rate in L/hour 

  Serial.print (Calc, DEC); //Prints the number calculated above
  Serial.print (" L/hour\r\n"); //Prints "L/hour" and returns a  new line 
}


void check_flow_alarm_delay ()
{
  Serial.println("enter CFAD");
  int Calc;    
  NbTopsFan = 0;   //Set NbTops to 0 ready for calculations
//  sei();      //Enables interrupts
  Alarm.delay(1000);   //Wait 1 second
//  cli();      //Disable interrupts
  Calc = (NbTopsFan * 60 / 5.5); //(Pulse frequency x 60) / 5.5Q, = flow rate in L/hour 

  Serial.print (Calc, DEC); //Prints the number calculated above
  Serial.print (" L/hour\r\n"); //Prints "L/hour" and returns a  new line 
}

Answer 1

delay()使用中断。禁用它们会干扰功能。