How to access a string value in a while loop, outside of a while loop, all while inside a method

Question

The program itself is supposed to take user input and then read it off in the setup of how a Mad Lib is made. The user types in a noun, verb, adjective, etc, and then its all printed to the console in the form of a story.

import java.util.Scanner;

public class CSFacts {

    //first method
    public static String Noun(int on)   {           //Start first user input, start with noun
        String noun1;
        Scanner input=new Scanner(System.in);
            while (on==1)   {
                System.out.println("Please enter a Noun.");     //prompt user to input a noun to be used in the main program
                noun1=input.nextLine(); 
                on=on-1;}
            while (on==0)   {
                noun1=input.nextLine(); }
            return noun1;
        //return the noun as a string to be used in the main program
    }
    public static void main(String[] args)  {
        Noun(1);
        System.out.println("Be kind to your "+Noun(0));
    }

The problem that I'm facing is that noun1 wont cooperate with me in the aspect of it will not carry though from one parameter to the other. I need noun1 to carry from on=1 to on=0 so that when the method is called for a second time it reads "Be kind to your (insert noun1)". I cant seem to figure out how to carry it over. This is only a section of my lab, but if I can figure out how to do this, I can implicate it into my other methods.

KSFT · Accepted Answer · 2015-03-05 01:35:45Z

1

You seem to have an infinite loop:

while (on==0)   {
    noun1=input.nextLine(); }

You call your method with Noun(1), so on is 1. Your first loop will run once, then end, setting on to 0. Your second loop will then run forever because it never changes on, but keeps going until it gets changed, which will never happen.

To answer your actual question, noun1 is defined inside the method Noun(), so it's a local variable and can't be used anywhere else. To fix that, you could define it as a static variable outside of Noun() like this:

static String noun1;

Another option would just be to use the value returned from Noun() inside your main method, instead of discarding it, like you're doing:

String noun1 = Noun(1);

answered Mar 5, 2015 at 1:35

KSFT

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5 Comments

Mark Wilcoxen Over a year ago

So i changed on to go back to 1 once its called a second time so it shouldn't be an infinite loop now. But the main problem now is that it keeps telling me on the return that noun1 may not be initialized, and i cant run the program to try and figure out whats wrong.

KSFT Over a year ago

@MarkWilcoxen What's the point of the loops? Why not just run each piece of code once?

Mark Wilcoxen Over a year ago

Awesome! Thank you so much KSFT! I ended up putting noun1 outside the method and the program runs just fine. Once again, thank you so much for your help, im sure my question was really basic/simple.

KSFT Over a year ago

@MarkWilcoxen If I helped, feel free to vote up and accept my answer to mark it as the one that worked for you.

KSFT Over a year ago

@MarkWilcoxen Oh yeah, you can't vote until you reach 15 reputation, but accepting my answer does help me.

erisco · Accepted Answer · 2015-03-05 02:29:51Z

There seems to be confusion in the control flow of your program. Remember that your program is a sequence of instructions. If you trace your control flow carefully, step by step, you will discover some problems. I will help detail what your program does and I will leave it to you to reason about it.

public static void main(String[] args)  {
    Noun(1);                                         // (0)
    System.out.println("Be kind to your "+Noun(0));  // (1)
}

I have added comments to label each statement. Your program always starts in the main method and on the first statement. Here is what main does.

(0) call Noun with on = 1, goto (1)
(1)
- call Noun with on = 0, store result in tmp1
- append the string stored in tmp1 to "Be kind to your ", store in tmp1
- print the string stored in tmp1

I added extra detail to (1) and invented the variable name tmp1 to ease the explanation — your program will store the data somewhere.

It is important to note that calling a function yields control to that function. This means the caller does not execute its next statement until the callee has executed all of its statements.

public static String Noun(int on)   {
    String noun1;                                   // (0)
    Scanner input = new Scanner(System.in);         // (1)
    while (on == 1) {                               // (2)
        System.out.println("Please enter a Noun."); // (3)
        noun1 = input.nextLine();                   // (4)
        on = on - 1;                                // (5)
    }
    while (on == 0) {                               // (6)
        noun1 = input.nextLine();                   // (7)
    }
    return noun1;                                   // (8)
}

Here is what the Noun function does.

(0) initialize noun1 to null, goto (1)
(1) construct a new Scanner object and store in input, goto (2)
(2) if on == 1 goto (3) else goto (6)
(3) print "Please enter a Noun.", goto (4)
(4) read a line of input and store into noun1, goto (5)
(5) store on - 1 into on, goto (2)
(6) if on == 0 goto (7) else goto (8)
(7) read a line of input and store into noun1, goto (6)
(8) halt and return the value of noun1

Notice there are a few problems just considering Noun by itself. I will focus on just one to help you get started.

If on is not equal to 0 or 1 then Noun will return null, which is why your IDE warns you. Why? To answer that, we just have to assume the hypothetical situation that on is neither 0 nor 1 and then manually execute the Noun function.

For simplicity I will choose on to be 3. I will keep track of the state of all variables as the function executes. To the left of the state I will indicate which statement I am executing, which I have decided only by following the program.

(0) on = 3, noun1 = null
(1) on = 3, noun1 = null, input = new Scanner(System.in)
(2) going to (6) because on is not equal to 1 on = 3, noun1 = null, input = new Scanner(System.in)
(6) going to (8) because on is not equal to 0 on = 3, noun1 = null, input = new Scanner(System.in)
(8) halting and returning value of noun1 on = 3, noun1 = null, input = new Scanner(System.in)

I suspect there is also confusion with variable scope. This indicates the lifetime of the variable. I will just briefly mention that variables declared in a function are destroyed (go out of scope) when the function halts (returns a value or throws an exception).

While this is a very clear and detailed explanation of how Java works, it doesn't seem to actually answer the question.

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