1 year ago
#347940
OptiShen
How do I use serial.available 2 times in this code?
int input;
int choices;
void setup() {
Serial.begin(9600);
Serial.println("BIN, DEC, HEX, OCT Converter\n");
retry:
Serial.println("Kindly choose a number system to convert (1, 2, 3, 4)");
Serial.println("1. BIN\n2. DEC\n3. HEX\n4. OCT\n");
while (Serial.available() == 0) {}
choices = Serial.parseInt();
if (choices == 1) {
Serial.println("You have chosen BIN, please enter a valid input");
}
else if (choices == 2) {
Serial.println("You have chosen DEC, please enter a valid input");
}
else if (choices == 3) {
Serial.println("You have choses HEX, please enter a valid input");
}
else if (choices == 4) {
Serial.println("You have chosen OCT, please enter a valid input");
}
else {
Serial.println("Invalid choice."); \
goto retry;
}
switch (choices) {
case 1: while (Serial.available() == 0) {}
input = Serial.parseInt();
Serial.print("You have input ");
Serial.print(input);
break;
case 2: while (Serial.available() == 0) {}
input = Serial.parseInt();
Serial.print("You have input ");
Serial.print(input);
break;
}
}
void loop() {
}
In the part switch
, when I run the code, it shows only 0 and it doesn't give me the option to enter at the serial monitor again.
I can't seem to figure this out, I'm a beginner who's trying to improve in the semantics of Arduino IDE. Thank you
arduino-ide
arduino-esp8266
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