Develop Reference

Can you replace ints in int Arrays?

I'm looking for a way to replace every int that is greater than 9 in an array with the sum of the digits. I'm gonna demonstrate an example below.
I have this array:
int[] ints2 = new int[] { ints[0] * 2, ints[1], ints[2] * 2, ints[3], ints[4] * 2, ints[5], ints[6] * 2, ints[7], ints[8] * 2, ints[9]};
And I want to replace the first int to whatever the sum is of the two digits.
Lets say ints[0] = 9. And ints[0]*2 = 18. The sum should be 1+8 = 9.
Can I use some kind of replace method for ints? Or do you guys have any better ideas of how to deal with this issue?
I tried it like this, but obviously I'm wrong:
foreach (int number in ints)
{
int newNumberInt;
if (number > 9)
{
string newNum = number.ToString();
char[] charNum = newNum.ToCharArray();
int[] intNum = Array.ConvertAll(charNum, n => (int)Char.GetNumericValue(n));
newNumberInt = intNum[0] + intNum[1];
}
newNumberInt = number;
}

You are close. Your mistakes are an incorrect addition of digits, and failure to replace the values in the array. Here is a one line method that adds all the digits in an integer (using System.Linq):
//Adds up all the digits in a number
private static int AddDigits(int number) => number.ToString().Sum(digitCharacter => int.Parse(digitCharacter.ToString()));
For replacing the values, a foreach loop directly on the array won't work, due to changing the contents of the array during enumeration. A simple solution is to use a for loop.
for (int i = 0; i < ints.Length; i++)
{
var value = ints[i];
ints[i] = value > 9 ? AddDigits(value) : value;
}
Note: the AddDigits function I wrote only works for positive integers

Incorrect values when converting char digits to int

My end goal is to take a number like 29, pull it apart and then add the two integers that result. So, if the number is 29, for example, the answer would be 2 + 9 = 11.
When I'm debugging, I can see that those values are being held, but it appears that other values are also being incorrect in this case 50, 57. So, my answer is 107. I have no idea where these values are coming from and I don't know where to begin to fix it.
My code is:
class Program
{
static void Main(string[] args)
{
int a = 29;
int answer = addTwoDigits(a);
Console.ReadLine();
}
public static int addTwoDigits(int n)
{
string number = n.ToString();
char[] a = number.ToCharArray();
int total = 0;
for (int i = 0; i < a.Length; i++)
{
total = total + a[i];
}
return total;
}
}

As mentioned the issue with your code is that characters have a ASCII code value when you cast to int which doesn't match with the various numerical digits. Instead of messing with strings and characters just use good old math instead.
public static int AddDigits(int n)
{
int total = 0;
while(n>0)
{
total += n % 10;
n /= 10;
}
return total;
}
Modulo by 10 will result in the least significant digit and because integer division truncates n /= 10 will truncate the least significant digit and eventually become 0 when you run out of digits.

Your code is actually additioning the decimal value of the char.
Take a look at https://www.cs.cmu.edu/~pattis/15-1XX/common/handouts/ascii.html
Decimal value of 2 and 9 are 50 and 57 respectively. You need to convert the char into a int before doing your addition.
int val = (int)Char.GetNumericValue(a[i]);

Try this:
public static int addTwoDigits(int n)
{
string number = n.ToString();
char[] a = number.ToCharArray();
int total = 0;
for (int i = 0; i < a.Length; i++)
{
total = total + (int)Char.GetNumericValue(a[i]);
}
return total;
}
Converted number to char always returns ASCII code.. So you can use GetNumericValue() method for getting value instead of ASCII code

Just for fun, I thought I'd see if I could do it in one line using LINQ and here it is:
public static int AddWithLinq(int n)
{
return n.ToString().Aggregate(0, (total, c) => total + int.Parse(c.ToString()));
}
I don't think it would be particularly "clean" code, but it may be educational at best!

You should you int.TryParse
int num;
if (int.TryParse(a[i].ToString(), out num))
{
total += num;
}

Your problem is that you're adding char values. Remember that the char is an integer value that represents a character in ASCII. When you are adding a[i] to total value, you're adding the int value that represents that char, the compiler automatic cast it.
The problem is in this code line:
total = total + a[i];
The code above is equal to this code line:
total += (int)a[i];
// If a[i] = '2', the character value of the ASCII table is 50.
// Then, (int)a[i] = 50.
To solve your problem, you must change that line by this:
total = (int)Char.GetNumericValue(a[i]);
// If a[i] = '2'.
// Then, (int)Char.GetNumericValue(int)a[i] = 2.
You can see this answer to see how to convert a numeric value
from char to int.
At this page you can see the ASCII table of values.

public static int addTwoDigits(int n)
{
string number = n.ToString()
char[] a = number.ToCharArray();
int total = 0;
for (int i = 0; i < a.Length; i++)
{
total += Convert.ToInt32(number[i].ToString());
}
return total;
}

You don't need to convert the number to a string to find the digits. #juharr already explained how you can calculate the digits and the total in a loop. The following is a recursive version :
int addDigit(int total,int n)
{
return (n<10) ? total + n
: addDigit(total += n % 10,n /= 10);
}
Which can be called with addDigit(0,234233433)and returns 27. If n is less than 10, we are counting the last digit. Otherwise extract the digit and add it to the total then divide by 10 and repeat.
One could get clever and use currying to get rid of the initial total :
int addDigits(int i)=>addDigit(0,i);
addDigits(234233433) also returns 27;
If the number is already a string, one could take advantage of the fact that a string can be treated as a Char array, and chars can be converted to ints implicitly :
var total = "234233433".Sum(c=>c-'0');
This can handle arbitrarily large strings, as long as the total doesn't exceed int.MaxValue, eg:
"99999999999999999999".Sum(x=>x-'0'); // 20 9s returns 180
Unless the number is already in string form though, this isn't efficient nor does it verify that the contents are an actual number.

Convert "int" into "char" inside char array

I am trying to do string manipulation. Here's my C# code :
static void Main(string[] args)
{
string input;
string output;
int length;
Console.WriteLine("input = ");
input = Console.ReadLine();
length = input.Length;
if ((input != "") || (length != 0))
{
Random randem = new Random();
int i = -1; //because I do not want the first number to be replaced by the random number
char[] characters = input.ToCharArray();
while (i < length)
{
int num = randem.Next(0, 9);
char num1 = Convert.ToChar(num);
i = i + 2; //so that every next character will be replaced by random number.. :D
characters[i] = num1; //*error* here
}
output = new string(characters);
Console.WriteLine(output);
}
For example:
User input : "i_love_to_eat_fish"
Desired output : "i2l4v1_9o5e8t7f8s2"
notice that the only unchanged character in
the char[] characters is : "i l v _ o e t f s". (desired output from the program)
I've already tried using this code, but still,
keep getting error at characters[i] = num1;
Am I on the right track?

I'm guessing the error you get is IndexOutOfRangeException this is because of the i = i + 2;. The while makes sure that i is less than length, but then adding 2 could result in it being more. Just add a check that it isn't beyond the length.
i = i + 2;
if(i < length)
characters[i] = num1;
Or just change to a for loop.
Random randem = new Random();
char[] characters = input.ToCharArray();
for(int i = 1; i < length; i += 2)
{
int num = randem.Next(1, 10); // max value is exclusive
char num1 = num.ToString()[0];
characters[i] = num1;
}
output = new string(characters);
Console.WriteLine(output);
Also as Shar1er80 points out you're currently converting the digit to the char that has the same ASCII value, and not the the actual characters that represent the digit. The digits 0-9 are represented by the the values 48-57. You can change the call to Random.Next to be:
int num = randem.Next(48, 58); // The upper bound is exclusive, not inclusive
char num1 = (char)num;
Or as Shar1er80 does it
int num = randem.Next(0,10) // Assumming you want digits 0-9
char num1 = num.ToString[0];
Also note that the max value for Random.Next is exclusive, so if you want to include the possibility of using a 9 you have to use an upper bound that is 1 greater than the greatest value you want.

Whenever you reach i = 17 you add 2 to i . That makes i = 19 with length of input equal to 18 that causes out of range exception.

The error you are getting is IndexOutOfTheRangeException, which explains everthing in itself.
It means that index you are feeding to array in the loop is going beyond its length-1 (as arrays have 0-based indexing)
So when you do i+2, you need to check if i+2 is not exceeding i.length-1 at any point of time; which does in your loop.
In general just check if you are supplying indexes between 0 and Array.Length-1

its because you start at index -1, and characters doesn't contain an index of -1.
EDIT: Sorry no the corrct answer is it must be while(i < length - 2)

Change this line
char num1 = Convert.ToChar(num);
To
char num1 = num.ToString()[0];
Then... Put
characters[i] = num1;
In an if block
if (i < length)
characters[i] = num1;

Converting an integer to an array of digits [duplicate]

This question already has answers here:
Is there an easy way to turn an int into an array of ints of each digit?
(11 answers)
Closed 7 years ago.
I want to know if there is a way in C# to convert an integer to an array of digits so that I can perform (Mathematical) operations on each digit alone.
Example: I need the user to input an integer i.e 123, 456
then the program creates two arrays of three elements {1,2,3}, {4,5,6}.

Off the top of my head:
int i = 123;
var digits = i.ToString().Select(t=>int.Parse(t.ToString())).ToArray();

You could create such array (or List) avoiding string operations as follows:
int x = 123;
List<int> digits = new List<int>();
while(x > 0)
{
int digit;
x = Math.DivRem(x, 10, out digit);
digits.Add(digit);
}
digits.Reverse();
Alternative without using the List and the List.Reverse:
int x = 456;
int[] digits = new int[1 + (int)Math.Log10(x)];
for (int i = digits.Length - 1; i >= 0; i--)
{
int digit;
x = Math.DivRem(x, 10, out digit);
digits[i] = digit;
}
And one more way using ToString:
int x = 123;
int[] digits = Array.ConvertAll(x.ToString("0").ToCharArray(), ch => ch - '0');

You can use this and not convert to a string:
var digits = new List<int>();
var integer = 123456;
while (integer > 0)
{
digits.Add(integer % 10);
integer /= 10;
}
digits.Reverse();

need a better way add leading digits to int and return array of digits

I need to create a modulus check which adds leading digits, lets say 0, to a seed int. I then need to return an array of digits in the array as I need to do a calculation on each digit to return a new whole number.
my code is as follows,
var seed = 1234;
var seedString = seed.ToString();
var test = new List<int>();
for(int i = 0; i < 10 - seedString.Length; i++)
{
test.Add(0);
}
var value = seed;
for(int i = 0; i < seedString.Length; i ++)
{
test.Insert(10 - seedString.Length, value % 10);
value = value / 10;
}
is there an easier way of doing this?

If you want to convert your number to a 10-digit string, you can format the number using a Custom Numeric Format String as follows:
string result = seed.ToString("0000000000");
// result == "0000001234"
See: The "0" Custom Specifier
If you need a 10-element array consisting of the individual digits, try this:
int[] result = new int[10];
for (int value = seed, i = result.Length; value != 0; value /= 10)
{
result[--i] = value % 10;
}
// result == new int[] { 0, 0, 0, 0, 0, 0, 1, 2, 3, 4 }
See also: Fastest way to separate the digits of an int into an array in .NET?

Try this:
int myNumber = 1234;
string myStringNumber = myNumber.ToString().PadLeft(10, '0');
HTH

Another shorthand way of getting the string representation would be
int num = 1234;
num.ToString("D10");