c++ - 在 Arduino 中存储上次读数

Question

你好，我正在尝试编写一个草图，从超声波距离传感器读取距离，并且仅在先前和当前读数不同(20 厘米)时才发送数据。下面是我的草图；

const int TRIG_PIN = 12;
const int ECHO_PIN = 11;

#include <SoftwareSerial.h>
SoftwareSerial BTserial(2, 3); // RX | TX
// Anything over 400 cm (23200 us pulse) is "out of range"
const unsigned int MAX_DIST = 23200;

void setup() {

  // The Trigger pin will tell the sensor to range find
  pinMode(TRIG_PIN, OUTPUT);
  digitalWrite(TRIG_PIN, LOW);

  // We'll use the serial monitor to view the sensor output
  Serial.begin(9600);
  BTserial.begin(9600); 
}

void loop() {

  unsigned long t1;
  unsigned long t2;
  unsigned long pulse_width;
  float cm;
  float inches;
  float lastcm;
  float diff;

  // Hold the trigger pin high for at least 10 us
  digitalWrite(TRIG_PIN, HIGH);
  delayMicroseconds(10);
  digitalWrite(TRIG_PIN, LOW);

  // Wait for pulse on echo pin
  while ( digitalRead(ECHO_PIN) == 0 );

  // Measure how long the echo pin was held high (pulse width)
  // Note: the () microscounter will overflow after ~70 min
  t1 = micros();
  while ( digitalRead(ECHO_PIN) == 1);
  t2 = micros();
  pulse_width = t2 - t1;

  // Calculate distance in centimeters and inches. The constants
  // are found in the datasheet, and calculated from the assumed speed 
  //of sound in air at sea level (~340 m/s).
  cm = pulse_width / 58.0;
  diff = cm - lastcm;
  lastcm = cm;

  // Print out results
  if ( pulse_width > MAX_DIST ) {
    BTserial.write("Out of range");
    lastcm = cm;
  } else {
    Serial.println(diff);
    Serial.println("Or act value");
    Serial.println(cm);
    lastcm = cm;
    if (abs(diff)>20) {
      BTserial.println(cm);
      }


  }

  // Wait at least 60ms before next measurement
  delay(150);
}

然而，这并没有正确计算两个值之间的差异 - 因为串行监视器刚刚返回

29.24
Or act value
29.24
29.31
Or act value
29.31
28.83
Or act value
28.83

有什么想法吗？

Answer 1

您的变量“lastcm”在循环函数内声明。你有两个选择。要么在循环外部声明它，要么在你所做的地方声明它，但将其设为静态(在任何一种情况下，你还必须处理它包含的初始无效读数)