Making a menu through serial

Question

I'm working on a project for a robot and I'm trying to make a menu that can run on start up where the user can calibrate the mid-point for the servos (so they're stationary) and the ldrs so it can tell between black lines and white paper. So, I've made the menu a switch case where it calls the calibration when user enters a value, the calibration routines are their own functions.

What I want to know how to do is how to make it so when the user is finished with a calibrator it will go back to the menu function and also how to get the menu to exit on user input. So far when the menu calls a function it will just loop on it.

The menu:

void Menu(){
 Serial.println("|****************************|");
 Serial.println("|**|Ardnoid Configuration |**|");
 Serial.println("|**| Menu |**|");
 Serial.println("|****************************|");
 Serial.println("");
 Serial.println("Select one of the following options:");
 Serial.println("1 LDR configuration");
 Serial.println("2 Servo calibration");
 Serial.println("3 Exit configuration menu");
 while(!Serial.available()){}
 int i = Serial.parseInt();
 switch(i){
 case 1:
 break;
 case 2:
 while(srvConOn == true){
 srvConfig();
 }
 break;strong text
 case 3:
 Serial.println("Leaving configuration menu"); 
 break;
 default:
 Serial.println("Unrecognized command, try again!");
 } 
 }

and the servo calibration:

void srvConfig(){
Serial.println("");
Serial.println("Begin servo stop calibration routine");
int x = 0;
 while(x == 0){
 Serial.println("Enter values for servos until both are stationary");
 delay(100);
 Serial.println("Enter value for left servo");
 while (!Serial.available()) {;}
 left = Serial.parseInt();
 Serial.println(left);
 delay(100);
 Serial.println("Enter value for right servo");
 while (!Serial.available()) {;}
 right = Serial.parseInt();
 Serial.println(right);
 delay(100);
 srvR.write(right);
 srvL.write(left);
 Serial.println("have both servos stopped?");
 Serial.println("1 = Yes | 2 = No"); 
 int j;
 while (!Serial.available()) {;}
 j = Serial.parseInt();
 if (j == 1) {
 Serial.println("Saving calibration");
 EEPROM.write(leftMid, left);
 EEPROM.write(rightMid, right);
 srvConOn == false; 
 x == 1;
 } else if (j == 2) {
 Serial.println("Try again");
 Serial.println();
 }
 }
}

Answer 1

You could put your switch/case inside an infinite loop, and return from the Menu() function when the user presses '3'. E.g.:

void Menu() {
 Serial.println(menu_banner);
 for (;;) {
 switch (Serial.read()) {
 case '1': ldrConfig(); break;
 case '2': srvConfig(); break;
 case '3': return;
 default: continue; // includes the case 'no input'
 }
 }
}

• ROff: Good work around for the menu! But how would you suggest getting from the functions back to the menu switch?
• Edgar Bonet: @ROff: This is already handled here: when srvConfig() returns, we break out of the switch, then we get into another iteration of the infinite for (;;) loop, then back to switch (Serial.read()). The infinite loop is only broken when the user types '3'.
• ROff: I just ran it through and it seems like whenever I try to save the srvConfig instead of returning to the menu it will just run srvConfig again
• Edgar Bonet: @ROff: In srvConfig(), replace x == 1 with x = 1, or more simply with return (and then get rid of x)..