Error 2 Use of unassigned local variable 'Y'

Question

Again i run into an error i don't mean to bug anyone but I'm getting an error on this code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Input_Program
{
    class Program
    {
       private static void Main()
        {

           char Y;
            char N;

           Console.WriteLine("Welcome to my bool program!");
           Console.WriteLine("Input a NON capital y or n when told to.");




            if(Y == 'y')
            {
                Console.WriteLine("Thank you,Please wait.....");
            }
        }
    }
}

Thanks for you answers!

The general rule is that you should always *initialize* a variable when you declare it. So you shouldn't write `char Y;`, but instead write `char Y = 'y';`. — Cody Gray - on strike, Mar 23 '11 at 03:21

score 1 · Accepted Answer · answered Mar 23 '11 at 03:14

1

Your variable char Y is not initialized before using. Try to give a default value when declaring.

EDIT It seems that you want the users to input something, and assign it to the variable Y. Try:

Y = Console.ReadKey().KeyChar;

answered Mar 23 '11 at 03:14

Cheng Chen

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But if it has a value then i can't do the if else's – unknown Mar 23 '11 at 03:16
`Console.ReadKey()` allows users input something, and assign it to the variable `y`, you can use `if-else` to do something according to the value of `y`. – Cheng Chen Mar 23 '11 at 03:21

score 1 · Answer 2 · answered Mar 23 '11 at 03:14

1

if(Y == 'y')

Y is a local variable which isn't assigned anything. So, you assign it any value before the if statement to make any comparison.

Y = 'a';  // or some character

answered Mar 23 '11 at 03:14

Mahesh

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If `Y` is assigned `a`, then `Y=='y'` yields to **false** and so the statements in `if` won't execute. – Mahesh Mar 23 '11 at 03:20

score 1 · Answer 3 · answered Mar 23 '11 at 03:16

1

You're not setting Y to anything, and you're also not reading anything from the keyboard.

answered Mar 23 '11 at 03:16

Andrew Cooper

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JTew · Answer 4 · 2011-03-23T03:29:15.943

1

You could explicitly set it to null.

char Y = '<whatever_is_the_default_char>';

That would get rid of the compiler error.

The root cause of the compiler error is that when it goes to compile the if conditional nothing as been assigned to Y. The above is considered an assignment.

edited Mar 23 '11 at 03:29

answered Mar 23 '11 at 03:21

JTew

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Error 2 Use of unassigned local variable 'Y'

4 Answers4

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