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C# Calling Random Variables

Posted on 2012-04-07
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Last Modified: 2012-04-07
Need a extra pair of eyes... What am I doing wrong with comparing random numbers and the input string


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

namespace test
{
    public class Program
    {
        static void Main(string[] args)
        {
            string inputString;
            string r, p, c;
            int randomNumber;
            string numberString;

            Random ranNumberGenerator = new Random();
            randomNumber = ranNumberGenerator.Next(4);


            Console.Write("Enter r, p, or c");
            inputString = Console.ReadLine();
            numberString = Convert.ToInt32(inputString);
            randomNumber =  + ranNumberGenerator.Next(4);    

           if (inputString == randomNumber)
               Console.WriteLine("Draw");
           else Console.WriteLine("Lose");
           Console.Read();
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Question by:December2000
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6 Comments
 
LVL 20

Expert Comment

by:BuggyCoder
ID: 37819820
randomNumber is an integer and inputString a string, here is the modified version:-

string inputString;
            string r, p, c;
            int randomNumber;
            int numberString;

            Random ranNumberGenerator = new Random();
            randomNumber = ranNumberGenerator.Next(4);


            Console.Write("Enter r, p, or c");
            inputString = Console.ReadLine();
            numberString = Convert.ToInt32(inputString);
            randomNumber = +ranNumberGenerator.Next(4);

            if (numberString == randomNumber)
                Console.WriteLine("Draw");
            else Console.WriteLine("Lose");
            

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0
 

Author Comment

by:December2000
ID: 37819980
Thank you!

I am getting an error on

 numberString = Convert.ToInt32(inputString);

say's "Input string was not in a correct format."
0
 
LVL 20

Expert Comment

by:BuggyCoder
ID: 37819986
you must be entering a string in console...
use int.tryparse to check if integer is added or not....
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Author Comment

by:December2000
ID: 37820030
Thanks,  as obvious I am a newbie, When I use number it works,  How do I use the int.tryparse? How would I compare r to random generated  r, p, s ? I am having a datatype issue
0
 
LVL 20

Accepted Solution

by:
BuggyCoder earned 2000 total points
ID: 37820045
Random class generates Random Number, here is a random character generator to help you:-

static class RandomLetter
    {
        static Random _random = new Random();
        public static char GetLetter()
        {
            // This method returns a random lowercase letter.
            // ... Between 'a' and 'z' inclusize.
            int num = _random.Next(0, 26); // Zero to 25
            char let = (char)('a' + num);
            return let;
        }
    }

Open in new window


Just Call RandomLetter.GetLetter() to get your random letter....

Reference:
http://www.dotnetperls.com/random-lowercase-letter

Then you can have your input string and compare it directly with the user input(assuming that user enters a single character only)
0
 

Author Closing Comment

by:December2000
ID: 37820056
Awesome still not there but, I am a lot further than I was... sorta stuck because the only characters that the computer should be able to choose from is r, p, s (it is a rock paper scissor game) ... However, you have more than earned the points already... thank you :)
0

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