Arduino map()方法 - 为什么？

Question

我只是看一些示例代码并遇到了一条线,我不完全理解为什么需要这样做.我知道你正在接受一个模拟值.这个值显然在0到1024之间？为什么是这样？为什么输出需要映射在0到255之间？是什么决定了这里使用的论点？有问题的一行:

   // map it to the range of the analog out:
      outputValue = map(sensorValue, 0, 1024, 0, 255); 

代码中突出显示:

created 29 Dec. 2008
 Modified 4 Sep 2010
 by Tom Igoe

 This example code is in the public domain.

 */

// These constants won't change.  They're used to give names
// to the pins used:
const int analogInPin = A0;  // Analog input pin that the potentiometer is attached to
const int analogOutPin = 9; // Analog output pin that the LED is attached to

int sensorValue = 0;        // value read from the pot
int outputValue = 0;        // value output to the PWM (analog out)

void setup() {
  // initialize serial communications at 9600 bps:
  Serial.begin(9600); 
}

void loop() {
  // read the analog in value:
  sensorValue = analogRead(analogInPin);            
  **// map it to the range of the analog out:
  outputValue = map(sensorValue, 0, 1024, 0, 255);**  
  // change the analog out value:
  analogWrite(analogOutPin, outputValue);           

  // print the results to the serial monitor:
  Serial.print("sensor = " );                       
  Serial.print(sensorValue);      
  Serial.print("\t output = ");      
  Serial.println(outputValue);   

  // wait 10 milliseconds before the next loop
  // for the analog-to-digital converter to settle
  // after the last reading:
  delay(10);                     
}

非常感谢回复.

Answer 1

模拟输出仅具有0到255之间的可接受范围.

因此,该值必须映射在可接受的范围内.

地图方法的文档在这里:http://arduino.cc/en/Reference/map

由于Arduino的analogRead分辨率为0-1023,模拟写入分辨率仅为0-255,因此电位器的原始数据需要在使用之前进行缩放...

这个解释来自Arduino传感器教程(在'Code'标题下):http: //arduino.cc/en/Tutorial/AnalogInOutSerial