Java duplicate local variable. Stumped

Question

I am brand new to coding and trying to get my second program working. It is pretty straight forward as to what it does, but it is throwing an error on line 24 "Duplicate local variable confirm". Can't quite work out why it doesn't like what I'm doing.

Scanner userInput = new Scanner(System.in);  
char confirm;

do{

System.out.println("Welcome to the story teller");
System.out.println("What is your name?");
String name = userInput.nextLine();

System.out.println("How old are you?");
int age = userInput.nextInt();

System.out.println("What country would you like to visit?");
String country = userInput.nextLine();

System.out.println("Great! So your name is" + name + ", you are" + age + "years old and you would like to visit" + country + "?");
System.out.println("Press Y to continue or N to start over");
char confirm = userInput.next().charAt(0);

  if (confirm !='y' || confirm !='n'){
 System.out.println("Sorry that input is not valid, please try again"); 
}
  else {
  System.out.println(name + "landed in" + country + "at the age of" + age + ".");
  }
} while(confirm == 'Y'|| confirm == 'y');

you declared char confirm two times within the same (local) scope. either remove the line char confirm; or turn the line char confirm = userInput.next().charAt(0); in confirm = userInput.next().charAt(0); I would recommend the second. — Stultuske, Mar 11 '19 at 07:39

score 1 · Answer 1 · answered Mar 11 '19 at 07:40

1

You're declaring confirm twice. Change the second declaration to just assigning to it and you should be OK:

confirm = userInput.next().charAt(0);
// No datatype, so you aren't declaring confirm, just assigning to it

answered Mar 11 '19 at 07:40

Mureinik

252,575
45
248
283

Thanks heaps! Would have taken forever to find that – Noob Mar 11 '19 at 07:45

score 1 · Answer 2 · answered Mar 11 '19 at 07:42

1

Because your "confirm" variable already defined in the scope (second row). If you want to assign a value, just write confirm = userInput.next().charAt(0);

answered Mar 11 '19 at 07:42

derUser

21
4

user7294900 · Answer 3 · 2019-03-11T07:55:18.957

0

Another option to fix is to remove the unnecessary declaration char confirm;

And use it only when needed

char confirm = userInput.next().charAt(0);

As @ScaryWombat suggested, you will need to change scope of the variable (currently while is in different scope than do )

edited Mar 11 '19 at 07:55

answered Mar 11 '19 at 07:45

user7294900

47,183
17
74
157

Thanks so much for your help! – Noob Mar 11 '19 at 07:46

Van_Cleff · Answer 4 · 2019-03-11T10:32:41.057

It seems apart from re-declaration of the variable confirm there are one or more issue -

Issue 1:

After int age = userInput.nextInt(). It won't prompt for country input and will prompt Press Y to continue or N to start over.

Cause of this issue:

Since you are using int age = userInput.nextInt(); the scanner will only take the integer value from the input and will skip the \n newline character.

Fix

As a workaround, I've added userInput.nextLine(); after int age = userInput.nextInt(); such that it will consume the \n character after nextInt().

Issue 2:

After the 1'st iteration, this line will cause issueconfirm = userInput.next().charAt(0);.

Cause of this issue:

In 2'nd iteration you won't get a prompt to enter the name as the line String name = userInput.nextLine(); will take \n from the last iteration as input and will skip and prompt for age How old are you?.

Fix

As a workaround, I've added userInput.nextLine(); after confirm = userInput.next().charAt(0); such that it will consume the \n character after userInput.next().charAt(0) and the next iteration will go as expected.

Issue 3:

This logic if (confirm !='y' || confirm !='n') expects only y and n in lowercase but here while(confirm == 'Y'|| confirm == 'y') you are expection y and Y both.

Fix - I've added the necessary changes in the code below but would recommend you do change it to a switch case.

NOTE:

It is not recommended to do userInput.nextLine() after every input and you could simply parse it. See here for further information.

I'm not recommending it but this will get you program working

Scanner userInput = new Scanner(System.in);
    char confirm;

    do {

        System.out.println("Welcome to the story teller");
        System.out.println("What is your name?");
        String name = userInput.nextLine();

        System.out.println("How old are you?");
        int age = userInput.nextInt();

        userInput.nextLine(); //adding this to retrieve the \n from nextint()

        System.out.println("What country would you like to visit?");
        String country = userInput.nextLine();

        System.out.println("Great! So your name is " + name + ", you are " + age
                + "years old and you would like to visit " + country + " ?");
        System.out.println("Press Y to continue or N to start over");
        confirm = userInput.next().charAt(0);

        userInput.nextLine(); //adding this to retrieve the \n this will help in next iteration

        System.out.println(name + " landed in " + country + " at the age of " + age + ".");

        if (confirm == 'y' || confirm == 'Y') {
            continue; // keep executing, won't break the loop
        } else if (confirm == 'n' || confirm == 'N') {
            break; // breaks the loop and program exits.
        } else {
            System.out.println("Sorry that input is not valid, please try again");
            // the program will exit
        }
    } while (confirm == 'Y' || confirm == 'y');
}

Recommending that you use switch case instead of if comparasion of confirmation and parse the character and integer input and remove the arbitary userInput.nextLine() added as workaround.

Java duplicate local variable. Stumped

4 Answers4