提高代码可重用性

构建一层抽象来操作Arduino一系列数字端口输出是很多人需要的，之前为了方便操作任意类型的流水灯，不管怎么跳转，只要把简单的逻辑整理好，操作一个10进制的变量即可。

如果说把Arduino当做黑盒子，来操作数字端口是常见的事情，下面这个例子讲告诉你构建简单抽象的方法与好处。

//to config the quantity of leds
const int ledNum = 8;
//use an integer array to set the Pin of leds
const int ledPin[] = { 2, 3, 4, 5, 6, 7, 8, 9 };
//use every bits of a decimal long interger record leds' state
long ledState = 10110001L;
 
//initializing function
void setup()
{
  setLedPin();
  Serial.begin(9600);
}
 
//set all the led ports OUTPUT
void setLedPin()
{
  for(int counter = 0; counter &lt; ledNum; counter ++)
  {
    pinMode(ledPin[counter], OUTPUT);
    digitalWrite(ledPin[counter], LOW);
  }
}
 
//this function will be excuted repeatedly
void loop()
{
  //detectButton();
  lightLed();
}
 
//light the leds
void lightLed()
{
  for(int counter = 0; counter &lt; ledNum; counter ++)
  {
    digitalWrite(ledPin[counter],
      true == fetchLedBit(ledState, counter)  ? HIGH : LOW); //translate HIGH/LOW to true/false
  }
}
 
//fetch the bitNum of temp, if its 1 return true, else return 0
boolean fetchLedBit(long temp, int bitNum)
{
  for(int counter = 0; counter &lt; bitNum; counter ++)
    temp /= 10;
  //ledState /= (long)pow(10, bitNum-1);
  if ( 1 == ( temp % 10 ))
    return true;
  else
    return false;
}
 
//if directionNum -1, ledState Shift right
//if directionNum 1, ledState Shift left
void shiftNum(int directionNum)
{
  if ( ledState != 1 || ledState != pow(2, ledNum) )
  {
    if( -1 == directionNum )
      ledState /= 10;
    else if( 1 == directionNum )
      ledState *= 10;
  }
}
 
//if left button is push, return 1
//if right button is push, return -1
int detectButton()
{
  return 0;
}