我写了一个小脚本来用风速计测量风速。 我想测量2秒,而不是测量8秒...
如果没有风,睡眠效果很好,但是在风速计旋转时,arduino不会进入睡眠状态。无论如何,我将如何修正代码以使其入睡。
这是我的代码。
#include "LowPower.h"
const byte interruptPin = 3; // anemomter input to digital pin
volatile unsigned long elapsedTime = 0;
int interval;
long WindAvr = 0; // sum of all wind speed between update
int measure_count = 0; // count each mesurement
unsigned int WindSpeed;
void setup() {
Serial.begin(9600);
attachInterrupt(
digitalPinToInterrupt(interruptPin), anemometerISR,
FALLING); // setup interrupt on anemometer input pin, interrupt will
// occur whenever falling edge is detected
}
void loop() {
WindAvr = 0;
measure_count = 0;
sei(); // Enables interrupts
delay(2000); // Wait x second to average
cli(); // Disable interrupts
LowPower.powerDown(SLEEP_8S, ADC_OFF, BOD_OFF);
WindSpeed = WindAvr / measure_count;
Serial.println(WindSpeed);
}
void anemometerISR() {
cli(); // Disable interrupts
static unsigned long previousTime = 0;
unsigned long time = millis();
if (time - previousTime > 15) { // debounce the switch contact.
elapsedTime = time - previousTime;
previousTime = time;
if (elapsedTime < 2000) {
interval = (22500 / elapsedTime) * 0.868976242;
WindAvr += interval; // add to sum of average wind values
++measure_count; // add +1 to counter
} else {
++measure_count; // add +1 to counter
}
}
sei(); // Enables interrupts
}
答案 0 :(得分:2)
即使使用I
未设置全局cli()
,外部中断(一旦被“取消屏蔽”)也能够唤醒设备。
更改
cli(); // Disable interrupts
LowPower.powerDown(SLEEP_8S, ADC_OFF, BOD_OFF);
到
detachInterrupt(digitalPinToInterrupt(interruptPin));
LowPower.powerDown(SLEEP_8S, ADC_OFF, BOD_OFF);
attachInterrupt(digitalPinToInterrupt(interruptPin), anemometerISR, FALLING);
也...我认为您不需要在ISR中使用cli()
和sei()
语句。进入/退出ISR时,全局中断允许位自动被禁用/使能。从Atmega644手册中:
发生中断时,将清除全局中断允许I位,并禁用所有中断。用户软件可以将逻辑1写入I位以启用嵌套中断。然后,所有允许的中断都可以中断当前的中断程序。当执行“从中断返回”指令RETI时,会自动设置I位。