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c# large fractional decimal split

My Requirement is Fractional Amount FIrst 2 decimal part add customer bank account and others fractional part add in dispute wallet account .
var amount = 40.235667745465465
I want to convert it 2 different variable
var customerBalance = // ??? - should be 40.23
var disputeBalance = amount - customerBalance
How can I do calculate the step marked ??? ?

This will work fine.
var firstAmount = Math.Floor(amount / 0.01) / 100 ;
var secondVariable = amount - firstAmount;

You probably want:
var firstAmount = Math.Round(amount, 2);
but note that this can round up as well as down; you may want to check whether secondVariable comes out negative, and if so: compensate.
Another way to look at it is to multiply by 100 and take the integer/decimal parts (hint: Math.Floor), then divide by 100 again.

If you are looking for a string, then it can be:
string secondVariableAsString = string.Format("{0:0.00}", secondVariable);
Another is:
Math.Truncate(100 * secondVariable) / 100;
However, this will cause overflow for large numbers.

Calculate number of distinct values between two numbers at a given precision

Context: I am building a random-number generating user interface where a user can enter values for the following:
lowerLimit: the lower limit for each randomly generated number
upperLimit: the upper limit for each randomly generated number
maxPrecision: the maximum precision each randomly generated number
Quantity: the maximum number of random number values to be generated
The question is: how can I ensure that at a given lowerLimit/upperLimit range and at a given precision, that the user does not request a greater quantity than is possible?
Example:
lowerLimit: 1
upperLimit: 1.01
maxPrecision: 3
Quantity: 50
At this precision level (3), there are 11 possible values between 1 and 1.01: 1.000, 1.001, 1.002, 1.003, 1.004, 1.005, 1.006, 1.007, 1.008, 1.009, 1.100, yet the user is asking for the top 50.
In one version of the function that returns only distinct values that match user criteria, I am using a dictionary object to store already-generated values and if the value already exists, try another random number until I have found X distinct random number values where X is the user-desired quantity. The problem is, my logic allows for a never-ending loop if the number of possible values is less than the user-entered quantity.
While I could probably employ logic to detect runaway condition, I thought it would be a nicer approach to somehow calculate the quantity of possible return values in advance to make sure it is possible. But that logic is eluding me. (Haven't tried anything because I can't think of how to do it).
Please note: I did see question Generating random, unique values C# but is does not address the specifics of my question relating to number of possible values at a given precision and subsequent runaway condition.
private Random RandomSeed = new Random();
public double GetRandomDouble(double lowerBounds, double upperBounds, int maxPrecision)
{
//Return a randomly-generated double between lowerBounds and upperBounds
//with maximum precision of maxPrecision
double x = (RandomSeed.NextDouble() * ((upperBounds - lowerBounds))) + lowerBounds;
return Math.Round(x, maxPrecision);
}
public double[] GetRandomDoublesUnique(double lowerBounds, double upperBounds, int maxPrecision, int quantity)
{
//This method returns an array of doubles containing randomly-generated numbers
//between user-entered lowerBounds and upperBounds with a maximum precision of
//maxPrecision. The array size is capped at user-entered quantity.
//Create Dictionary to store number values already generated so we can ensure
//we don't have duplicates
Dictionary<double, int> myDoubles = new Dictionary<double, int>();
double[] returnValues = new double[quantity];
double nextValue;
for (int i = 0; i < quantity; i++)
{
nextValue = GetRandomDouble(lowerBounds, upperBounds, maxPrecision);
if (!myDoubles.ContainsKey(nextValue))
{
myDoubles.Add(nextValue, i);
returnValues[i] = nextValue;
}
else
{
i -= 1;
}
}
return returnValues;
}

Number of items can be computed by just subtracting "position" of first from last (pseudo-code below, use Math.Pow to compute 10^x):
(int)(last * 10 ^ precision) - (int)(first * 10 ^ precision)
This may need to be adjusted depending on whether you want boundaries and whether you take decimal (precise) or float/double as input - some +/-1 and Math.Round may need to be sprinkled in to get desired results for all expected values.
After you get number of items there are essentially two cases
there are significantly more choices that desired results (i.e. 1 to 100, take 5 random numbers) - use code you have to filter out duplicates.
there the number of choices is close or less than desired number of results (i.e. 1 to 10, return 11 random numbers) - pre-generate the list of all value and shuffle.
Experiment with the boundary between "significantly more" and "close" - I'd use 25% as boundary ( i.e. 1 to 100, take 76 - use shuffling) to avoid excessive retires close to the end (which is exact reason of slowness/infinite retries of basic approach).
Correct implementation of shuffle is in Randomize a List<T> (check out similar posts like Generating random, unique values C# for more discussion).

The easiest way would probably be to convert the values to integers by multiplying them by 10 ^ precision and then subtract
int lowerInt = (int)(lower * (decimal)Math.Pow(10, precision));
int higherInt = (int)(higher * (decimal)Math.Pow(10, precision));
int possibleValues = higherInt - lowerInt + 1
I feel like it would defeat the purpose of you project to require the user to know how many possible values there are in advance, since it seems like thats what they are hitting this function for in the first place. I'm assuming that requirement was just to alleviate the technical issues you were having. You can just change your loop to this now
for (int i = 0; i < possibleValues; i++)

This is what worked based on Josh Williard's answer.
public double[] GetRandomDoublesUnique(double lowerBounds, double upperBounds, int maxPrecision, int quantity)
{
if (lowerBounds >= upperBounds)
{
throw new Exception("Error in GetRandomDoublesUnique is: LowerBounds is greater than UpperBounds!");
}
//These next few lines are for the purpose of determining the maximum possible number of return values
//possibleValues is populated to prevent a runaway condition that could occurs if the
//max possible values--at the given precision level--is less than the user-selected quantity.
//i.e. if user selects 1 to 1.01, precision of 3, and quantity of 50, there would be a problem
// if we didn't limit loop to the 11 possible values at precision of 3:
//1.000, 1.001, 1.002, 1.003, 1.004, 1.005, 1.006, 1.007, 1.008, 1.009, 1.010
int lowerInt = (int)(lowerBounds * (double)Math.Pow(10, maxPrecision));
int higherInt = (int)(upperBounds * (double)Math.Pow(10, maxPrecision));
int possibleValues = higherInt - lowerInt + 1;
//Create Dictionary to store number values already generated so we can ensure
//we don't have duplicates
Dictionary<double, int> myDoubles = new Dictionary<double, int>();
double[] returnValues = new double[(quantity>possibleValues?possibleValues:quantity)];
double NextValue;
//Iterate through and generate values--limiting to both the user-selected quantity and # of possible values
for (int i = 0; (i < quantity)&&(i<possibleValues); i++)
{
NextValue = GetRandomDouble(lowerBounds, upperBounds, maxPrecision);
if (!myDoubles.ContainsKey(NextValue))
{
myDoubles.Add(NextValue, i);
returnValues[i] = NextValue;
}
else
{
i -= 1;
}
}
return returnValues;
}

How to take digits from two different numbers and form a new one

I have the following problem here:My input is several lines of 2 digit numbers and I need to make a new number using the second digit of the first number and the first of the next one.
Example:
int linesOfNumbers = Convert.ToInt32(Console.ReadLine());
for(int i = 0,i<linesOfNumbers,i++)
{
int numbers = Conver.ToInt32(Console.ReadLine());
//that's for reading the input
}
I know how to separate the numbers into digits.My question is how to merge them.
For example if your input is 12 and 21 the output should be 22.

I like oRole's answer, but I think they're missing a couple things with the example input that you provided in your comment. I'll also point out some of the errors in the code that you have.
First off, if you're only given the input 12,23,34,45, then you don't need to call Console.ReadLine within your for loop. You've already gotten the input, you don't need to get any more (from what you've described).
Secondly, unless you're doing mathematical operations, there is no need to store numerical data as ints, keep it as a string, especially in this case. (What I mean is that you don't store Zip Codes in a database as a number, you store it as a string.)
Now, onto the code. You had the right way to get your data:
var listOfNumbers = Console.ReadLine();
At that point, listOfNumbers is equal to "12,23,34,45". If you iterate on that variable as a string, you'll be taking each individual character, including the commas. To get each of the numbers to operate on, you'll need to use string.Split.
var numbers = listOfNumbers.Split(',');
This turns that list into four different two character numbers (in string form). Now, you can iterate over them, but you don't need to worry about converting them to numbers as you're operating on the characters in each string. Also, you'll need a results collection to put everything into.
var results = new List<string>();
// Instead of the regular "i < numbers.Length", we want to skip the last.
for (var i = 0; i < numbers.Length - 1; i++)
{
var first = numbers[i];
var second = numbers[i + 1]; // This is why we skip the last.
results.Add(first[1] + second[0]);
}
Now your results is a collection of the numbers "22", "33", and "44". To get those back into a single string, you can use the helper method string.Join.
Console.WriteLine(string.Join(",", results));

You could use the string-method .Substring(..) to achieve what you want.
If you want to keep int-conversion in combination with user input, you could do:
int numA = 23;
int numB = 34;
int resultAB = Convert.ToInt16(numA.ToString().Substring(1, 1) + numB.ToString().Substring(0, 1));
Another option would be to take the users input as string values and to convert them afterwards like that:
string numC = "12";
string numD = "21";
int resultCD = Convert.ToInt16(numC.Substring(1, 1) + numD.Substring(0, 1));

I hope this code snippet will help you combining your numbers. The modulo operator (%) means: 53 / 10 = 5 Rest 3
This example shows the computation of the numbers 34 and 12
int firstNumber = 34 - (34 % 10) // firstNumber = 30
int secondNumber = 12 % 10; // secondNumber = 2
int combined = firstNumber + secondNumber; // combined = 32
EDIT (added reading and ouput code):
boolean reading = true;
List<int> numbers = new ArrayList();
while(reading)
{
try
{
int number = Convert.ToInt32(Console.ReadLine());
if (number > 9 && number < 100) numbers.Add(number);
else reading = false; // leave reading process if no 2-digit-number
}
catch (Exception ex)
{
// leave reading process by typing a character instead of a number;
reading = false;
}
}
if (numbers.Count() > 1)
{
List<int> combined = new ArrayList();
for (int i = 1; i <= numbers.Count(); i++)
{
combined.Add((numbers[i-1] % 10) + (numbers[i] - (numbers[i] % 10)));
}
//Logging output:
foreach (int combination in combined) Console.WriteLine(combination);
}

As you mention, if you already have both numbers, and they are always valid two digit integers, following code should work for you.
var num1 = 12;
var num2 = 22;
var result = (num2 / 10)*10 + (num1 % 10);
num2/10 returns the first digit of second number, and num1 % 10 returns the second digit of the first number.

The % and / signs are your savior.
If you want the 'ones' digit of a number (lets call it X), simply do X%10 - the remainder will be whatever number is in the 'ones' digit. (23%10=3)
If, instead, the number is two digits and you want the 'tens' digit, divide it by ten. (19/10=1).
To merge them, multiply the number you want to be in the 'tens' digit by ten, and add the other number to it (2*10+2=22)

There are other solutions like substring, etc and many one have already given it above. I am giving the solution VIA LINQ, note that this isn't efficient and it's recommended only for learning purpose here
int numA = 12;
int numB = 21 ;
string secondPartofNumA = numA.ToString().Select(q => new string(q,1)).ToArray()[1]; // first digit
string firstPartofNumB = numB.ToString().Select(q => new string(q,1)).ToArray()[0]; // second digit
string resultAsString = secondPartofNumA + firstPartofNumB;
int resultAsInt = Convert.ToInt32(resultAsString);
Console.WriteLine(resultAsString);
Console.WriteLine(resultAsInt);

How to get count of numbers in int and how to split a number without making a string

I have a number like 601511616
If all number's length is multiple of 3, how can a split the number into an array without making a string
Also, how can I count numbers in the int without making a string?
Edit: Is there a way to simply split the number, knowing it's always in a multiple of 3... good output should look like this: {616,511,601}

You can use i % 10 in order to get the last digit of integer.
Then, you can use division by 10 for removing the last digit.
1234567 % 10 = 7
1234567 / 10 = 123456
Here is the code sample:
int value = 601511616;
List<int> digits = new List<int>();
while (value > 0)
{
digits.Add(value % 10);
value /= 10;
}
// digits is [6,1,6,1,1,5,1,0,6] now
digits.Reverse(); // Values has been inserted from least significant to the most
// digits is [6,0,1,5,1,1,6,1,6] now
Console.WriteLine("Count of digits: {0}", digits.Count); // Outputs "9"
for (int i = 0; i < digits.Count; i++) // Outputs "601,511,616"
{
Console.Write("{0}", digits[i]);
if (i > 0 && i % 3 == 0) Console.Write(","); // Insert comma after every 3 digits
}
IDEOne working demonstration of List and division approach.
Actually, if you don't need to split it up but only need to output in 3-digit groups, then there is a very convenient and proper way to do this with formatting.
It will work as well :)
int value = 601511616;
Console.WriteLine("{0:N0}", value); // 601,511,616
Console.WriteLine("{0:N2}", value); // 601,511,616.00
IDEOne working demonstration of formatting approach.

I can't understand your question regarding how to split a number into an array without making a string - sorry. But I can understand the question about getting the count of numbers in an int.
Here's your answer to that question.
Math.Floor(Math.Log10(601511616) + 1) = 9
Edit:
Here's the answer to your first question..
var n = 601511616;
var nArray = new int[3];
for (int i = 0, numMod = n; i < 3; numMod /= 1000, i++)
nArray[i] = numMod%1000;
Please keep in mind there's no safety in this operation.
Edit#3
Still not perfect, but a better example.
var n = 601511616;
var nLength = (int)Math.Floor(Math.Log10(n) + 1)/ 3;
var nArray = new int[nLength];
for (int i = 0, numMod = n; i < nLength; numMod /= 1000, i++)
nArray[i] = numMod%1000;
Edit#3:
IDEOne example http://ideone.com/SSz3Ni
the output is exactly as the edit approved by the poster suggested.
{ 616, 511, 601 }

Using Log10 to calculate the number of digits is easy, but it involves floating-point operations which is very slow and sometimes incorrect due to rounding errors. You can use this way without calculating the value size first. It doesn't care if the number of digits is a multiple of 3 or not.
int value = 601511616;
List<int> list = new List<int>();
while (value > 0) // main part to split the number
{
int t = value % 1000;
value /= 1000;
list.Add(t);
}
// Convert back to an array only if it's necessary, otherwise use List<T> directly
int[] splitted = list.ToArray();
This will store the splitted numbers in reverse order, i.e. 601511616 will become {616, 511, 601}. If you want the numbers in original order, simply iterate the array backwards. Alternatively use Array.Reverse or a Stack

Since you already know they are in multiples of 3, you can just use the extracting each digit method but use 1000 instead of 10. Here is the example
a = 601511616
b = []
while(a):
b.append(a%1000)
a = a//1000
print(b)
#[616, 511, 601]

First and last two numbers of an integer windows phone

How to get the first two numbers of an integer and the last two in c#.
For example:
int complete = 1935;
how to make,
int firtTwo = 19;
and
int secondTwo = 35;
Please help.

Is it always a four-digit positive value? If so, I'd go for:
int hundreds = complete / 100;
int tensAndUnits = complete % 100;
If it's any arbitrary number, so you'd want "935" to be "93" and "35" you might as well use string operations:
string text = complete.ToString();
int first = int.Parse(text.Substring(0, 2));
int last = int.Parse(text.Substring(text.Length - 2, 2));
(Note that this will blow up for single-digit values...)
It would help if you could say what this is meant to be for though.