Why do I have to create a new Scanner?

I'm have just begun learning java and have been asked to create a program which:

Assigns 2 Arrays: Suit, Rank
Uses the Math.random class to determine the cards drawn. (I realize that this isn't a very realistic approach to how cards are drawn.)
I ran into troubles when I tried to add the option for the user to choose to draw a new hand.
For some reason the compiler will run, but when I go to ask for a new hand I get the message: "static error: undefined name 'yes' "

My Code before fix

Code java:

---

import java.util.Scanner;
 
public class exercise5{
  public static void main(String[] args){
    Scanner input = new Scanner(System.in);
    String again;
 
    do{
      System.out.println("How many cards would you like?");
      int counter = input.nextInt();
 
      String[] suit = {"Diamonds", "Clubs", "Hearts", "Spades"};
      String[] rank = {"Ace", "King", "Queen", "Jack", "10", "9", "8", "7", "6", "5", "4", "3", "2", "1"};
 
      while (counter > 0){
 
        double rand1 = Math.random() * 4;
        int ran1 = (int)rand1;
        double rand2 = Math.random() * 14;
        int ran2 = (int)rand2;
 
        if (rand1 >= 3.9 || rand2 >= 13.9)
          System.out.println("You got a joker!!!");
 
        else
          System.out.println(rank[ran2]+ " of " + suit[ran1]);
 
      counter --;
      }
 
 
    System.out.println("Would you like a new hand? ");
    again = input.nextLine();
 
    }while (again.startsWith("y") || again.startsWith("Y"));
  }
}

---

I have found that assigning a new Scanner just before my do-while condition will fix my program. I am looking for a reason as to why. I'm not sure I understand what is happening. If anyone could help me out, it would be greatly appreciated.


My Code After Fix

Code java:

---

import java.util.Scanner;
 
public class exercise5{
  public static void main(String[] args){
    Scanner input = new Scanner(System.in);
    String again;
 
    do{
      System.out.println("How many cards would you like?");
      int counter = input.nextInt();
 
      String[] suit = {"Diamonds", "Clubs", "Hearts", "Spades"};
      String[] rank = {"Ace", "King", "Queen", "Jack", "10", "9", "8", "7", "6", "5", "4", "3", "2", "1"};
 
      while (counter > 0){
 
        double rand1 = Math.random() * 4;
        int ran1 = (int)rand1;
        double rand2 = Math.random() * 14;
        int ran2 = (int)rand2;
 
        if (rand1 >= 3.9 || rand2 >= 13.9)
          System.out.println("You got a joker!!!");
 
        else
          System.out.println(rank[ran2]+ " of " + suit[ran1]);
 
      counter --;
      }
 
 
    Scanner in = new Scanner(System.in);
    System.out.println("Would you like a new hand? ");
    again = in.nextLine();
 
    }while (again.startsWith("y") || again.startsWith("Y"));
  }
}

---

I don't think the compiler message comes from the code you posted (which does not contain the symbol 'yes' referred to in the message).

Both version of the code compile as they are.

Quote:

---

Originally Posted by Currahe

I'm have just begun learning java and have been asked to create a program which:

Assigns 2 Arrays: Suit, Rank
Uses the Math.random class to determine the cards drawn. (I realize that this isn't a very realistic approach to how cards are drawn.)
I ran into troubles when I tried to add the option for the user to choose to draw a new hand.
For some reason the compiler will run, but when I go to ask for a new hand I get the message: "static error: undefined name 'yes' "

My Code before fix
.
.
.

---

That's funny. (I mean "funny-peculiar," not "funny-haha.")

When I run your first example, it asks "How many cards..." and when I enter an integer, it deals the cards and then quits without asking for anything else.

Here's the deal (Is it possible for you to pardon the pun? Sometimes I just can't help myself.)

When the user types a bunch of stuff and then presses 'Enter' all of the characters are in a system buffer and a special line termination character is appended. (A "newline" character, if you will.)

Then when the program executes the nextInt() method, the scanner builds an integer from characters in the system buffer, and stops scanning when it sees the 'newline' character. The 'newline' character is still in the system buffer at this point.

If the user had entered some integer digits and a space and some other stuff, all of the other stuff would still be in the input buffer..

Anyhow...

When the program executes the nextLine() method, the Scanner starts with whatever is left in the buffer from any previous calls to nextInt(), and the Scanner stops when it sees a 'newline' character. The effect is that if the user enters an integer and the program executes nextInt(), the nextLine() method sees the 'newline' character, removes it from the buffer and returns with an empty String. (If the input buffer is empty when a nextLine() is executed, the Scanner will wait for user input and then retrieve characters until it sees the 'newline' character. Note that the 'newline' character encountered by nextLine() will be removed from the input buffer, but will not appear in the String.

It's easy to avoid that by simply putting a "dummy" nextLine() immediately after the nextInt() instruction so that the nextLine() will see an empty buffer and will wait for further input.

So...

Maybe you can go back to the first example code in your post and make the following change:

Code java:

---

      System.out.println("How many cards would you like?");
      int counter = input.nextInt();
      input.nextLine(); // "Dummy input flushes the input buffer and throws away any detritus

---

Much cleaner than creating a brand new Scanner, right?


Cheers!

Z

Oh, well maybe it is just the compiler I am using... are you using Dr. Java?

Quote:

---

are you using Dr. Java?

---

No, I just compiled with javac.

Quote:

---

Originally Posted by Currahe

Oh, well maybe it is just the compiler I am using... are you using Dr. Java?

---

If I am the one you are asking, the answer is no.

For testing I used command line compile with javac and command line execution with java (OpenJDK Java version 1.6.0_22) on a Centos 5.8 Linux workstation.



Did you try your original code with the "dummy" nextLine() added, as I suggested/recommended?


Cheers!

Z

Thank you, i am very grateful for your explanation. Your solution is very useful as well, I will remember to keep it in mind. Like I said, I'm new to programming, and my teacher wasn't able to answer this for me, so thank god for this forum. lol Thanks again.