C 在10个led显示屏上显示healthbar

对于编程课程，我必须制作一个带有fubarino和10个led显示屏的小游戏。在游戏中的某一点上，我需要向玩家显示“healthbar”或分数。运行状况可以介于10和0之间。我用了这段代码

  value = -1;
//set 1 bit off per point lost
  for(int i = 10; i>points[playerNr]; i--){
    value &= !(1<<(points[playerNr]-1));
  }
  showValue(value);

value=-1；
//每丢失一点设置1位
对于（int i=10；i>点[playerr]；i--）{
值&=！（1您可以使用
，直接创建一个设置了“n个最低位”的整数，而不是循环清除位（1u抱歉，您能否详细说明一下，您的评论对我来说不是很清楚。您会怎么做（1u
u
表示未签名的
onliterals@Typhaon这不是一个变量，这是一个无符号后缀，用于使
1
具有类型
unsigned int
，而不是
int
。您可以使用
showValue（（1u So value=0；value |=（1）替换整个
for
循环
// playerNr is either 0 or 1 for one player or 2 for both
void showPoints(int playerNr){
  bool both = false;
  //subtract one point from both players if playerNr == 2
  if(playerNr == 2){
    both = true;
    playerNr = 1;
  }
  points[playerNr]--;
  value = -1;
  //flash all lights to show that the points are going to be shown
  showValue(value);
  delay(50);
  //set 1 bit off per point lost
  for(int i = 10; i>points[playerNr]; i--){
    value &= !(1<<(points[playerNr]-1));
  }
  showValue(value);
  //game position is either 0 or 9
  pos = 9*playerNr;
  //ascend is true if playerNr is 0 and vice versa
  ascend = !playerNr;
  if(both){
    pointLost = 0;
  } else {
    pointLost = -1;
  }
  //reset the game if points are 0
  if(points[playerNr]==0){
    points[0] = 10;
    points[1] = 10;
    points[playerNr]++;
    pointLost = playerNr;
  }
  timer = 1000;
}