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原来想实现的功能是有6个按钮开关,按Leda的时候向电脑发送一个字符,leda+ledb的时候发送另外的一个字符,依次类推。不按的时候不发送任何字符,但 这段代码跑起来的时候 总是不停的循环发送字符,那位帮我看看逻辑上哪里有问题,谢谢!
const int leda =2;
const int ledb =3;
const int ledc =4;
const int ledd =5;
const int lede =6;
const int ledf =7;
int valuea = 0;
int valueb = 0;
int valuec = 0;
int valued = 0;
int valuee = 0;
int valuef= 0;
void setup()
{
Serial.begin(9600);
pinMode(leda, INPUT);
pinMode(ledb, INPUT);
pinMode(ledc, INPUT);
pinMode(ledd, INPUT);
pinMode(lede, INPUT);
pinMode(ledf, INPUT);
}
void loop()
{
valuea = digitalRead(leda);
valueb = digitalRead(ledb);
valuec = digitalRead(ledc);
valued = digitalRead(ledd);
valuee = digitalRead(lede);
valuef = digitalRead(ledf);
if( valuea=valueb=valuec=valued=valuee=valuef==HIGH) { Serial.println("5"); delay(500);} else {Serial.println(""); delay(500); valuea=valueb=valuec=valued=valuee=valuef=LOW; }
if( valuea=valueb=valuec=valued=valuee==HIGH) { Serial.println("6"); delay(500);} else {Serial.println(""); delay(500); valuea=valueb=valuec=valued=valuee=LOW; }
if( valuea=valueb=valuec=valued==HIGH) { Serial.println("7"); delay(500);} else {Serial.println(""); delay(500); valuea=valueb=valuec=valued=LOW; }
if( valuea=valueb=valuec==HIGH) { Serial.println("1"); delay(500);} else {Serial.println(""); delay(500); valuea=valueb=valuec=LOW; }
if( valuea=valueb==HIGH) { Serial.println("2"); delay(500);} else {Serial.println(""); delay(500); valuea=valueb=LOW; }
if( valuea==HIGH) { Serial.println("3"); delay(500);} else {Serial.println(""); delay(500); valuea=LOW; }
if( valueb=valuec==HIGH) { Serial.println("4"); delay(500);} else {Serial.println(""); delay(500); valueb=valuec=LOW; }
}
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