Here's my code:
const int AnalogInPin1 = A1;
const int AnalogInPin2 = A2;
int SerialPrint = 0;
int SerialMonitor = 0;
void setup() {
Serial.begin(9600);
}
void loop() {
SerialPrint = analogRead(AnalogInPin1);
Serial.print("Sensor = ");
Serial.println(SerialPrint);
if (SerialPrint > 400) {
digitalWrite(3, LOW);
digitalWrite(4, HIGH);
delay(500);
}
else{
digitalWrite(4, HIGH);
digitalWrite(3, LOW);
delay(500);
}
}
void setup1() {
Serial.begin();
}
void loop1() {
SerialMonitor = analogRead(AnalogInPin2);
Serial.print("Sensor = ");
Serial.println(SerialMonitor);
if (SerialMonitor > 400) {
digitalWrite(6, LOW);
digitalWrite(5, HIGH);
delay(500);
}
else{
digitalWrite(5, HIGH);
digitalWrite(6, LOW);
delay(500);
}
}
I am trying to make a car guidance system like the ones found in underground parking lots. and I am trying to have two serial monitors read from 2 sensors.
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The characters above are looping in the serial monitor whilst using 19200 baud but not in 9600 baud why is that?
CodePudding user response:
If your question is about how to display two analogRead values in one SerialMonitor sketch, this should compile and work (if your SerialMonitor is set to 9600, of course):
const int AnalogInPin1 = A1;
const int AnalogInPin2 = A2;
int value1 = 0;
int value2 = 0;
void setup() {
Serial.begin(9600);
pinMode(3, OUTPUT);
pinMode(4, OUTPUT);
pinMode(5, OUTPUT);
pinMode(6, OUTPUT);
}
void loop() {
value1 = analogRead(AnalogInPin1);
Serial.print("Sensor1 = ");
Serial.print(value1);
if (value1 > 400) {
digitalWrite(3, LOW);
digitalWrite(4, HIGH);
}
else{
digitalWrite(4, HIGH);
digitalWrite(3, LOW);
}
value2 = analogRead(AnalogInPin2);
Serial.print("\t Sensor2 = ");
Serial.println(value2);
if (value2 > 400) {
digitalWrite(6, LOW);
digitalWrite(5, HIGH);
}
else{
digitalWrite(5, HIGH);
digitalWrite(6, LOW);
}
delay(500);
}
I left most of your code as is, just made it compile and work. If this is not what you want, please edit your question.