Looping through a switch statement

Question

So I am writing a menu based program, and I am stuck in one part. Here's my code:

public static void main(String [] args) throws FileNotFoundException {
    switch (menu()) {
        case 1:
            System.out.println("Stub 1");
            menu();
            break;
        case 2:
            System.out.println("Stub 2");
            menu();
            break;
        case 3:
            System.out.println("Stub 3");
            menu();
            break;
        case 4:
            System.out.println("Program Terminated");
            break;

    }

}

public static int menu() {
    System.out.println("Choose a task number from the following: ");
    System.out.println("\t1. - See histogram of name's popularity");
    System.out.println("\t2. - Compare two names in a specific decade");
    System.out.println("\t3. - Show what name had a specific rank for a certain decade");
    System.out.println("\t4. - Exit program");
    int opt = 0;
    int option = getInt(input,"Enter number (1-4): ", 1, 4);
    if (option == 1) {
        opt = 1;
    }
    else if (option == 2) {
        opt = 2;
    }
    else if (option == 3) {
        opt = 3;
    } 
    else {
        opt = 4;
    }
    return opt;
}

My question is, how can I get the menu to 'reset' after an option is pressed. For example, I choose 1, the program performs the action and after it's done, it shows the options menu again until I press 4 to terminate it.

The getInt method in my code simply returns an int between 1 and 4.

try removing menu() call from inside each case sub block. Four is the only one that doesn't call it again, that's probably why it breaks then. — Mark Giaconia, Nov 22 '14 at 02:14
Maybe this link can help you: http://stackoverflow.com/a/17015039/2444047 — Splamy, Nov 22 '14 at 02:17

score 2 · Accepted Answer · answered Nov 22 '14 at 02:22

One easy option is to declare a Boolean variable and wrap the switch in a while loop e.g.

Boolean quit = false;
while (!quit)        //or do-while
{
    int opt = menu();
    switch(opt)
    {
        //other cases...
        case 4:
            quit = true;
    }
}

I'm not sure why you're calling menu in each case though.

score 1 · Answer 2 · answered Nov 22 '14 at 02:17

1

I don't code in java but try pointing it back to a default case at the end of each case so that when you program finishes your action it will default back to menu.

answered Nov 22 '14 at 02:17

Twinnaz

44
10

score 1 · Answer 3 · answered Nov 22 '14 at 02:18

1

For my menus I always wrap the menu options and request in a do-while loop.

do{
menu code...
} while (menu() != 4);

answered Nov 22 '14 at 02:18

Spencer

453
4
21

This kinda works, but when it runs once, it skips the action the second time it runs, and so on – Zero Nov 22 '14 at 02:21
My bad, neglected to realize this was java before posting. Been coding in C++ all semester for school. – Spencer Nov 22 '14 at 02:27

Johny · Answer 4 · 2014-11-22T02:44:54.290

You can keep the code in an infinite loop and exit the program when 4 is pressed. It is not required to call menu() in all your cases because you have to display the menu only one in every iteration.

For making an infinite loop use

while(true) {
  //some code
}

for exiting the program use:

System.exit(0);

Try this :

while(true) {
     int choice = menu();
     switch (choice) {
        case 1:
            System.out.println("Stub 1");

            break;
        case 2:
            System.out.println("Stub 2");

            break;
        case 3:
            System.out.println("Stub 3");

            break;
        case 4:
            System.out.println("Program Terminated");
            System.exit(0); // for terminating the program

    }

}

Looping through a switch statement

4 Answers4