CODEWITHSUNDEEP
c#arrays
Prendoza
Prendoza

Reputation: 367

array values confusion

While working on a basic coding challenge I have come across a confusing situation that i am unable to fathom the cause of (I am new to programming)

whilst attempting to split a number into its individual digits, my int array contains the value 13 but returns the value 49??

there is probably an obvious reason for this and if so i apologize.

I have found an alternate way to split my numerical string into individual digits but would still like to know what i was doing wrong

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AddTheDigits
{
    class Program
    {
        static void Main(string[] args)
        {
            using (StreamReader reader = new StreamReader("TextFile1.txt")) 
            {
                    //int total = 0;
                    string line = reader.ReadLine();

                    //Console.WriteLine(line); //testing

                    char[] charArray = line.ToCharArray(0,1);

                    int[] intArray = new int[charArray.Length];

                    for (int i = 0; i < intArray.Length; i++)
                    {
                        intArray[i] = Convert.ToInt32(charArray[i]);
                        Console.Write(intArray[i]);
                    }
                }
            Console.ReadLine();
        }
    }
}

Upvotes: 1

Views: 83

Answers (4)

Amit
Amit

Reputation: 46323

you're taking you line (Which probably contains '13'), and grabbing the first char in it (line.ToCharArray(0,1);). You then convert this char (which has the numerical value 49, equal to '1' in UTF-16) to an int (still equals 49) and you print it. That's it. the '3' is left out due to the ToCharArray only grabbing the first character.

Upvotes: 2

Ortiga
Ortiga

Reputation: 8814

49 is ASCII code for 1.

Instead of using Int32.Convert, you should use Int32.Parse.

Upvotes: 0

Jonesopolis
Jonesopolis

Reputation: 25370

What you are doing is essentially:

"13".ToCharArray(0,1);

which will give you the same value as :

char c = '1';

which is 49, the ASCII code for '1'

Upvotes: 0

Krumelur
Krumelur

Reputation: 32497

You are probably mixing up the ASCII code for the digits with the digits themselves. Try replacing

intArray[i] = Convert.ToInt32(charArray[i]);

with 

intArray[i] = charArray[i] - '0';

and see if that helps.

Upvotes: 2

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