好的,所以我是Arduino编码的新手。我正在尝试创建一个程序,该程序将根据电位计的模拟输入打开一系列LED。这是我写的代码:
#define red 4
#define blue 3
#define yellow 2
void setup() {
pinMode(red, OUTPUT);
pinMode(blue, OUTPUT);
pinMode(yellow, OUTPUT);
}
void loop() {
int val = digitalRead(A0);
if val <= 341(digitalWrite(red, HIGH)
if val >= 682(digitalWrite(red, HIGH); digitalWrite(blue, HIGH);
if val >= 1023(digitalWrite(red, HIGH); digitalWrite(blue, HIGH); digitalWrite(yellow, HIGH);)
}
我不断收到错误代码"预期'('在'val'之前''",有谁知道这意味着什么以及如何解决它?
if 中的 C 语法不正确。这是操作方法:
if(val <= 341){
digitalWrite(red, HIGH);
}
if(val >= 682 && val < 1023){
digitalWrite(red, HIGH);
digitalWrite(blue, HIGH);
}
if(val >= 1023){
digitalWrite(red, HIGH);
digitalWrite(blue, HIGH);
digitalWrite(yellow, HIGH);
}