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C# Remove all preceding zeros from a double (no strings) [closed]

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I need a method to return the firsts non zero numbers from a double in the following way: Any number >= 1 or == 0 will return the same; All the rest will return as per the following examples:
(Please note that I am using double because the potential imprecision is irrelevant in the use case whereas saving memory is relevant).
double NumberA = 123.2; // Returns 123.2
double NumberB = 1.2; // Returns 1.2
double NumberC = 0.000034; // Returns 3.4
double NumberD = 0.3; // Returns 3.0
double NumberE = -0.00000087; // Returns -8.7

One option would be to iteratively multiply by 10 until you get a number greater than 1:
public double RemoveLeadingZeros(double num)
{
if (num == 0) return 0;
while(Math.Abs(num) < 1) { num *= 10};
return num;
}
a more direct, but less intuitive, way using logarithms:
public double RemoveLeadingZeros(double num)
{
if (num == 0) return 0;
if (Math.Abs(num) < 1) {
double pow = Math.Floor(Math.Log10(num));
double scale = Math.Pow(10, -pow);
num = num * scale;
}
return num;
}
it's the same idea, but multiplying by a power of 10 rather then multiplying several times.
Note that double arithmetic is not always precise; you may end up with something like 3.40000000001 or 3.3999999999. If you want consistent decimal representation then you can use decimal instead, or string manipulation.

I would start with converting to a string. Something like this:
string doubleString = NumberC.ToString();
I don't know exactly what this will output, you will have to check. But if for example it is "0.000034" you can easily manipulate the string to meet your needs.

C# Cannot convert `char' expression to type `string' [closed]

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I have this code:
public class Kata
{
public static bool Narcissistic(int value)
{
//add variale to hold final result
int finalResult = 0;
//get the length of the value input
int valLength = value.ToString().Length;
//convert value given into an array of ints
string valString = value.ToString();
//iterate over each number and multiply that number by the length of value input
for (int i=0; i < valLength; i++) {
//convert char at index[i] of stringified value to int and mutiply by # of digits, store to result
finalResult += int.Parse(valString[i]) * valLength;
}
//return the result
return finalResult == value;
}
}
I'm getting an error when I run this that I understand, but don't quite know how to fix. My goal is to take a number (i.e 1234) and multiply each number by the total number of digits it contains (i.e 1*4 + 2*4 + 3*4...etc).

Assuming you have just Latin digit characters 0 to 9, you could use (valString[i] - '0') instead of int.Parse(valString[i]).
valString[i] is a char, not a string (which int.Parse() would expect). And chars are automatically converted to their Unicode integer code in C#, and as thhe digits 0 to 9 have consecutive Unicode values, '0' - '0' would be the Unicode value of the character for the digit 0 minus itself, i. e. 0, and so on up to '9' - '0' being 9 more than 0, i. e. 9.
This code would not work if there could be non Latin digits or non digit characters in your string. And looking at your complete code, the assumption should be met, as actually, you are converting from an int.

Here's a better option that avoids strings altogether
public static bool Narcissistic(int value)
{
if(value < 0) return false;
int sum = 0;
int count = 0;
while(value>0)
{
sum += value%10;
value /= 10;
count++;
}
return value == (sum*count);
}
Basically the value%10 will give you the least significate digit. Then the value /= 10; will truncate that digit. Once value is zero you've seen all the digits. And your formula (4*1 + 4*2 + 4*3 + 4*4) can of course just be 4*(1+2+3+4), which is the sum of the digits times the number of digits.
And presumably no negative number would be "Narcissistic" so you can just return false for them.

valString[i] will be a char - because that's what indexing does to a string. int.Parse expects a string. The simplest thing to do would probably just
finalResult += int.Parse(valString[i].ToString()) * valLength;
Assuming to turn a single character digit into a one-digit integer is your intention.
This is less "dangerous" than the traditional hack of subtracting '0' because you can rely on int.Parse to throw an exception if the char is anything but numeric.
You will also need to make the input argument uint, or define a behavior for negative numbers.

int.Parse() expects a string, but valString[i] is a character. You could just change this to valString[i].ToString(), but rather than parsing you can take advantage of the fact that the digit characters are encoded in sequence by casting them to int and subtracting the value of the 0 integer:
public static bool Narcissistic(int value)
{
int zero = (int)'0';
string valueString = value.ToString();
int total = 0;
foreach(char c in valueString)
{
int digit = ((int)c - zero);
total += digit * valueString.Length;
}
return total == value;
}
or
public static bool Narcissistic(int value)
{
int zero = (int)'0';
string valueString = value.ToString();
return valueString.Select(c => (c - zero) * valueString.Length).Sum() == value;
}
If you really want to you can write the second option as a one-liner:
public static bool Narcissistic(int value)
{
return value.ToString().Select(c => c - '0').Sum() * value.ToString().Length == value;
}

Find Double/Float value is Odd or Even in C# [closed]

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How to find if double value is even or odd without converting them to int in C#? E.g.
123.0d - odd
456.0d - even
3.1415926d - floating point (neither odd nor even)

Try modulo % operator:
double x = 123;
string result = x % 2 == 0
? "even" : x % 2 == 1 || x % 2 == -1
? "odd"
: "floating point"; // e.g. 123.456789
Edit: When does it work? Floating point value (single, double) when it doesn't contain exponential part represents the (integer) value exactly.
https://en.wikipedia.org/wiki/Single-precision_floating-point_format
https://en.wikipedia.org/wiki/Double-precision_floating-point_format
So we have that the solution will work whenever x in these ranges
[-2**24..2**24] == [-16777216..16777216] (float)
[-2**52..2**52] == [-4503599627370496..4503599627370496] (double)
Please, notice, that since .Net assumes that negative % positive == nonpostive, e.g. -3 % 2 == -1 we have to check x % 2 == -1 as well as x % 2 == 1

well i think you answered your question by your own but here is a short code i wrote:
static void Main(string[] args)
{
double number=10.0;
if(number%2==0)
{
Console.WriteLine("Your number is even!");
}
else
{
Console.WriteLine("Your number is odd!");
}
Console.ReadLine();
}
this way you see really good if it works or not

public static bool IsEven(double x)
{
return x % 2 == 0;
}
Please note that it takes more code to make a helper method that does this than it does to actually do it, so if you ever put this in a production environment expect your QA team to send you nasty e-mails.

public bool isEven(double value)
{
decimal decVal = System.Convert.ToDecimal(value);
if (decVal % 2.0 == 0)
{
return true;
}
else
{
return false;
}
}
EDIT: if you convert to a decimal first it should work.

Algorithm: how many indivisible units are needed to overshadow a random number? [closed]

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How can I find how many indivisible units of 1,000 are needed to overshadow a random number?
For example, for random number 5,123 I'm going to need 6 x 1,000 to overshadow it, so: MyAlgorithm(5123, 1000) = 6
Edit1: I am sorry if despite my endeavor to articulate my problem into a meaningful description my dyslexia took over, I hope this edit makes it a bit more comprehensible.

Well, if I understand your question, it sounds like you could simply convert the parameters to decimals, divide, then use Math.Ceiling:
int output = (int)Math.Ceiling((decimal)5123 / (decimal)1000); // 6
Alternatively, you could avoid the conversions and rely purely on integer division and the % operator (modulus), like this:
int output = (5123 / 1000) + (5123 % 1000 == 0 ? 0 : 1);
If you want this in a method simply wrap it up like this:
static int MyAlgorithm(int a, int b)
{
return (a / b) + (a % b == 0 ? 0 : 1);
}

if I've understood you correctly, this is really just a one-liner:
public static int MyAlgorithm(int input, int units)
{
return input%units == 0 ? input/units : input/units + 1;
}
the only case when it isn't simply the result of input/units + 1 is the case when there is no remainder

public int MyAlgorithm(int x, int y)
{
int result = x / y;
return (result < 0) ? (result - 1): (result + 1);
}

Retrieving a value that is stored as MB [closed]

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I have a table that stores the amount of RAM a server has in a biginit column with values such as 2470208.
But how I can apply a data annotation or other validations to show only 2 instead of s470208. ?
I mean to always divide by 1 million and get the number on the left side of the digit ?

1) Use this for automatic thousands-unit:
string GetByteString(long n) {
int k=0;
string u=" kMGTEP";
while(n>1024) {
n>>=10;
k++;
}
return n.ToString() + u[k];
}
Call:
string s= GetByteString(1234567890123);
Debug.WriteLine(s);
2) But if you simply always want MB just shift by 20:
long n = 123456789;
string MB = (n>>20).ToString();
But this will show 0 if n goes below 1MB.
Reason:
1 kB = 2^10 = 1<<10 = 1024;
1 MB = 2^20 = 1<<20 = 1024*1024 = 1048576;
1 GB = 2^30 = 1<<30 = 1024*1024*1024 = 1073741824;

You tagged C# but mentioned a bigint column so it isn't clear whether you're looking for a database or C# solution. The following C# method will take the number of bytes as an integer and return a formatted string...
public string FormattedBytes(long bytes)
{
string units = " kMGT";
double logBase = Math.Log((double)bytes, 1024.0);
double floorBase = Math.Floor(logBase);
return String.Format("{0:N2}{1}b",
Math.Pow(1024.0, logBase - floorBase),
units.Substring((int)floorBase, 1));
}