Java中的逻辑和循环问题

Question

I started coding small program in Java. I wanted to exercise try-catch block, but i did not even come to that part and got stuck on loop part. I know this is very basic loop issue, but i guess i caught myself in a very simple logical problem. What I need from this program is if a user press 1, then to jump to switch statement and execute proper case. If a user press anything but 1 or 2, to go back to the MenuLoop function and execute it again until pressed correct number (1 or 2). I used While loop for control. Here is the code.

import java.util.Scanner;

    public class TryCatchExercise {

    public static void MenuLoop() {
    Scanner input = new Scanner(System.in);
    int choice;

    System.out.println("1. Check for Number 1");
    System.out.println("2. Check for Number 2");
    System.out.print("Please enter your choice... ");
    choice = input.nextInt();
    while (choice != 1 || choice != 2) {
        System.out.println("Invalid entry, press 1 or 2");
        MenuLoop();
    }    //Isn't it logical at this point for loop to be skipped and 
         // go to Switch if a user pressed 1 or 2.??

    switch (choice) {
    case 1:
        System.out.println("Pressed 1");
        break;
    case 2:
        System.out.println("Pressed 2");
        break;
    default:
        System.out.println("Invalid number");
    }
}

public static void main(String[] args) {
    MenuLoop();

}

}
 OUTPUT
 1. Check for Number 1
 2. Check for Number 2
 Please enter your choice... 1
 Invalid entry, press 1 or 2
 1. Check for Number 1
 2. Check for Number 2
 Please enter your choice... 2
 Invalid entry, press 1 or 2
 1. Check for Number 1
 2. Check for Number 2
 Please enter your choice... 5
 Invalid entry, press 1 or 2
 1. Check for Number 1
 2. Check for Number 2
 Please enter your choice... 

Answer 1

You need a logical and (not or) here

while (choice != 1 || choice != 2) {
    System.out.println("Invalid entry, press 1 or 2");
    MenuLoop();
} 

should be something like

while (choice != 1 && choice != 2) {
    System.out.println("Invalid entry, press 1 or 2");
    MenuLoop();
} 

or (using De Morgan's laws ) like

while (!(choice == 1 || choice == 2)) {
    System.out.println("Invalid entry, press 1 or 2");
    MenuLoop();
} 

Answer 2

Part of the problem is that you are recursively calling menuLoop from within menuLoop

If you had a while loop within main then you could just do a return if the proper keys is not pressed.

so main would be something like

while (!menuLoop () {
  System.out.println("Invalid entry, press 1 or 2");
}

and menuLoop would return a boolean

public static boolean MenuLoop() {

  ....

System.out.println("1. Check for Number 1");
System.out.println("2. Check for Number 2");
System.out.print("Please enter your choice... ");
choice = input.nextInt();
if(choice != 1 && choice != 2) {   // and NOT or
    return false;
}    

switch (choice) {
case 1:
    System.out.println("Pressed 1");
    break;
case 2:
    System.out.println("Pressed 2");
    break;
default:
    System.out.println("Invalid number");
}
return true;

Also please remember that the scanner.nextInt will not swallow up the Enter that may or may not be pressed.

Answer 3

Maybe you can try the following code. In your code , it's not need to use iteration.

choice = input.nextInt();
while (choice != 1 && choice != 2) {
    System.out.println("Invalid entry, press 1 or 2");
    choice = input.nextInt();
}   

Answer 4

This is a logical problem, the "choice" value should be either 1 or 2. In your (while) statement you are checking that "choice" is not different from 1 but also that "choice" is not different from 2. This condition is never reached because "choice" can be either 1 or 2 but not both values at the same time. This is only an explanation of the situation.