I have a problem with my unit converter app. I`m trying to prevent users from using any other characters than numbers but I need to allowed them to use "." How can I do this?
I spend like 2 days on that with no luck. Please help. Here is my code.
public string[] YardsToMeters(string yards)
{
if(yards == null || yards.Contains(" "))
{
string[] str = new string[1];
str[0] = "Please enter some values";
return str;
}
Here I`m checking for allowed characters:
else if((Regex.IsMatch(yards, #"^\d+$") == true) || (yards.Contains(".") && yards[0] != '.'
|| (yards.Contains('\n')) && (Regex.IsMatch(yards, #"^\d+$") == true) && !yards.Contains(".")))
{
if (!yards.Contains('\n') && Regex.IsMatch(yards, #"^\d+$") == true)
{
double d = double.Parse(yards, CultureInfo.InvariantCulture) * 0.9144f;
string[] array = new string[1];
array[0] = d.ToString();
return array;
}
else if(Regex.IsMatch(yards, #"^\d+$") == true || yards.Contains(".") && yards[0] != '.')
{
double[] doubles = new double[yards.Split("\r\n").Count()];
for (int i = 0; i < yards.Split("\r\n").Count(); i++)
{
if (yards.Contains("."))
{
double value = Convert.ToDouble(yards.Split("\r\n")[i].Replace('.',','));
doubles[i] += value;
string.Format("{0}", value * 0.9144f);
}
else
{
double value = Convert.ToDouble(yards.Split("\r\n")[i]);
doubles[i] += value;
string.Format("{0}", value * 0.9144f);
}
}
string[] strings = new string[doubles.Length];
for (int i = 0; i < yards.Split('\n').Length; i++)
{
strings[i] = string.Format("{0}", doubles[i] * 0.9144f);
}
return strings;
}
else
{
string[] str = new string[1];
str[0] = "Please use numbers only!";
return str;
}
}
else
{
string[] str = new string[1];
str[0] = "Please use numbers only! Enter is also not allowed";
return str;
}
}
Please help me with that.
I think you've reached a point where you've lost sight of the goal and you've just been endlessly modifying and tweaking a bad solution to try and cover all the cases where it falls down
Yards to Metres -> Multiply yards by 0.9144f
public double YardsToMetres(double yards){ //not sure why you're using string everywhere
return yards * 0.9144f;
}
public void YardsTextbox_TextChanged(object sender, EventArgs e){
try{
double y = double.Parse(_yardsTextbox.Text);
MessageBox.Show("In metres:" + YardsToMetres(y));
} catch(FormatException ex){
MessageBox.Show("The value you entered (" + _yardsTextbox.Text + ") cannot be converted to a number. Enter a number");
}
}
This is an example, knowing nothing about your UI.. It's a recommendation that you split off the conversion code from the rest of the code. It should take a number and return a number (don't make everything string just because the input/output supply/require it).
Your input routine that collects data from the user should parse their input into a number ready for the converter. If it fails to parse, ask them to correct it. Don't get too deep and meaningful/complex over this - just try to parse it, and if it fails, tell the user it failed - they'll work out what they did wrong. You can do SOME limited amount of help, like trimming whitespace off the start if you want (actually double.parse ignores whitespace so it's unnecessary) but i'd perhaps stop short of stripping out all non numbers etc, because youre then doing large amount of manipulation of th data the user entered, doing calcs on it and the user has no idea (unless you tell them) what data you ended up doing your calcs on.
If they write 0xFF80, and expect their hex number (equivalent of 65,408) to be converted to mtres, they might be surprised if you've stripped out the xFF, and then done the calc on 80
Similarly, double parse can be made to be sensitive to different cultures. Some people write a 1,234.5 as 1.234,5. If you decide to strip out commas because theyre thousand separators, you end up ruining someone's decimal number if they use comma as the decimal separator. Consider carefully how deep you want to get into checking and helping the user form a number correctly; you're better to just try and work with what they gave you and if it failed, tell them
How to convert a double into a floating-point string representation without scientific notation in the .NET Framework?
"Small" samples (effective numbers may be of any size, such as 1.5E200 or 1e-200) :
3248971234698200000000000000000000000000000000
0.00000000000000000000000000000000000023897356978234562
None of the standard number formats are like this, and a custom format also doesn't seem to allow having an open number of digits after the decimal separator.
This is not a duplicate of How to convert double to string without the power to 10 representation (E-05) because the answers given there do not solve the issue at hand. The accepted solution in this question was to use a fixed point (such as 20 digits), which is not what I want. A fixed point formatting and trimming the redundant 0 doesn't solve the issue either because the max width for fixed width is 99 characters.
Note: the solution has to deal correctly with custom number formats (e.g. other decimal separator, depending on culture information).
Edit: The question is really only about displaing aforementioned numbers. I'm aware of how floating point numbers work and what numbers can be used and computed with them.
For a general-purpose¹ solution you need to preserve 339 places:
doubleValue.ToString("0." + new string('#', 339))
The maximum number of non-zero decimal digits is 16. 15 are on the right side of the decimal point. The exponent can move those 15 digits a maximum of 324 places to the right. (See the range and precision.)
It works for double.Epsilon, double.MinValue, double.MaxValue, and anything in between.
The performance will be much greater than the regex/string manipulation solutions since all formatting and string work is done in one pass by unmanaged CLR code. Also, the code is much simpler to prove correct.
For ease of use and even better performance, make it a constant:
public static class FormatStrings
{
public const string DoubleFixedPoint = "0.###################################################################################################################################################################################################################################################################################################################################################";
}
¹ Update: I mistakenly said that this was also a lossless solution. In fact it is not, since ToString does its normal display rounding for all formats except r. Live example. Thanks, #Loathing! Please see Lothing’s answer if you need the ability to roundtrip in fixed point notation (i.e, if you’re using .ToString("r") today).
I had a similar problem and this worked for me:
doubleValue.ToString("F99").TrimEnd('0')
F99 may be overkill, but you get the idea.
This is a string parsing solution where the source number (double) is converted into a string and parsed into its constituent components. It is then reassembled by rules into the full-length numeric representation. It also accounts for locale as requested.
Update: The tests of the conversions only include single-digit whole numbers, which is the norm, but the algorithm also works for something like: 239483.340901e-20
using System;
using System.Text;
using System.Globalization;
using System.Threading;
public class MyClass
{
public static void Main()
{
Console.WriteLine(ToLongString(1.23e-2));
Console.WriteLine(ToLongString(1.234e-5)); // 0.00010234
Console.WriteLine(ToLongString(1.2345E-10)); // 0.00000001002345
Console.WriteLine(ToLongString(1.23456E-20)); // 0.00000000000000000100023456
Console.WriteLine(ToLongString(5E-20));
Console.WriteLine("");
Console.WriteLine(ToLongString(1.23E+2)); // 123
Console.WriteLine(ToLongString(1.234e5)); // 1023400
Console.WriteLine(ToLongString(1.2345E10)); // 1002345000000
Console.WriteLine(ToLongString(-7.576E-05)); // -0.00007576
Console.WriteLine(ToLongString(1.23456e20));
Console.WriteLine(ToLongString(5e+20));
Console.WriteLine("");
Console.WriteLine(ToLongString(9.1093822E-31)); // mass of an electron
Console.WriteLine(ToLongString(5.9736e24)); // mass of the earth
Console.ReadLine();
}
private static string ToLongString(double input)
{
string strOrig = input.ToString();
string str = strOrig.ToUpper();
// if string representation was collapsed from scientific notation, just return it:
if (!str.Contains("E")) return strOrig;
bool negativeNumber = false;
if (str[0] == '-')
{
str = str.Remove(0, 1);
negativeNumber = true;
}
string sep = Thread.CurrentThread.CurrentCulture.NumberFormat.NumberDecimalSeparator;
char decSeparator = sep.ToCharArray()[0];
string[] exponentParts = str.Split('E');
string[] decimalParts = exponentParts[0].Split(decSeparator);
// fix missing decimal point:
if (decimalParts.Length==1) decimalParts = new string[]{exponentParts[0],"0"};
int exponentValue = int.Parse(exponentParts[1]);
string newNumber = decimalParts[0] + decimalParts[1];
string result;
if (exponentValue > 0)
{
result =
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
}
else // negative exponent
{
result =
"0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Length) +
newNumber;
result = result.TrimEnd('0');
}
if (negativeNumber)
result = "-" + result;
return result;
}
private static string GetZeros(int zeroCount)
{
if (zeroCount < 0)
zeroCount = Math.Abs(zeroCount);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < zeroCount; i++) sb.Append("0");
return sb.ToString();
}
}
You could cast the double to decimal and then do ToString().
(0.000000005).ToString() // 5E-09
((decimal)(0.000000005)).ToString() // 0,000000005
I haven't done performance testing which is faster, casting from 64-bit double to 128-bit decimal or a format string of over 300 chars. Oh, and there might possibly be overflow errors during conversion, but if your values fit a decimal this should work fine.
Update: The casting seems to be a lot faster. Using a prepared format string as given in the other answer, formatting a million times takes 2.3 seconds and casting only 0.19 seconds. Repeatable. That's 10x faster. Now it's only about the value range.
This is what I've got so far, seems to work, but maybe someone has a better solution:
private static readonly Regex rxScientific = new Regex(#"^(?<sign>-?)(?<head>\d+)(\.(?<tail>\d*?)0*)?E(?<exponent>[+\-]\d+)$", RegexOptions.IgnoreCase|RegexOptions.ExplicitCapture|RegexOptions.CultureInvariant);
public static string ToFloatingPointString(double value) {
return ToFloatingPointString(value, NumberFormatInfo.CurrentInfo);
}
public static string ToFloatingPointString(double value, NumberFormatInfo formatInfo) {
string result = value.ToString("r", NumberFormatInfo.InvariantInfo);
Match match = rxScientific.Match(result);
if (match.Success) {
Debug.WriteLine("Found scientific format: {0} => [{1}] [{2}] [{3}] [{4}]", result, match.Groups["sign"], match.Groups["head"], match.Groups["tail"], match.Groups["exponent"]);
int exponent = int.Parse(match.Groups["exponent"].Value, NumberStyles.Integer, NumberFormatInfo.InvariantInfo);
StringBuilder builder = new StringBuilder(result.Length+Math.Abs(exponent));
builder.Append(match.Groups["sign"].Value);
if (exponent >= 0) {
builder.Append(match.Groups["head"].Value);
string tail = match.Groups["tail"].Value;
if (exponent < tail.Length) {
builder.Append(tail, 0, exponent);
builder.Append(formatInfo.NumberDecimalSeparator);
builder.Append(tail, exponent, tail.Length-exponent);
} else {
builder.Append(tail);
builder.Append('0', exponent-tail.Length);
}
} else {
builder.Append('0');
builder.Append(formatInfo.NumberDecimalSeparator);
builder.Append('0', (-exponent)-1);
builder.Append(match.Groups["head"].Value);
builder.Append(match.Groups["tail"].Value);
}
result = builder.ToString();
}
return result;
}
// test code
double x = 1.0;
for (int i = 0; i < 200; i++) {
x /= 10;
}
Console.WriteLine(x);
Console.WriteLine(ToFloatingPointString(x));
The problem using #.###...### or F99 is that it doesn't preserve precision at the ending decimal places, e.g:
String t1 = (0.0001/7).ToString("0." + new string('#', 339)); // 0.0000142857142857143
String t2 = (0.0001/7).ToString("r"); // 1.4285714285714287E-05
The problem with DecimalConverter.cs is that it is slow. This code is the same idea as Sasik's answer, but twice as fast. Unit test method at bottom.
public static class RoundTrip {
private static String[] zeros = new String[1000];
static RoundTrip() {
for (int i = 0; i < zeros.Length; i++) {
zeros[i] = new String('0', i);
}
}
private static String ToRoundTrip(double value) {
String str = value.ToString("r");
int x = str.IndexOf('E');
if (x < 0) return str;
int x1 = x + 1;
String exp = str.Substring(x1, str.Length - x1);
int e = int.Parse(exp);
String s = null;
int numDecimals = 0;
if (value < 0) {
int len = x - 3;
if (e >= 0) {
if (len > 0) {
s = str.Substring(0, 2) + str.Substring(3, len);
numDecimals = len;
}
else
s = str.Substring(0, 2);
}
else {
// remove the leading minus sign
if (len > 0) {
s = str.Substring(1, 1) + str.Substring(3, len);
numDecimals = len;
}
else
s = str.Substring(1, 1);
}
}
else {
int len = x - 2;
if (len > 0) {
s = str[0] + str.Substring(2, len);
numDecimals = len;
}
else
s = str[0].ToString();
}
if (e >= 0) {
e = e - numDecimals;
String z = (e < zeros.Length ? zeros[e] : new String('0', e));
s = s + z;
}
else {
e = (-e - 1);
String z = (e < zeros.Length ? zeros[e] : new String('0', e));
if (value < 0)
s = "-0." + z + s;
else
s = "0." + z + s;
}
return s;
}
private static void RoundTripUnitTest() {
StringBuilder sb33 = new StringBuilder();
double[] values = new [] { 123450000000000000.0, 1.0 / 7, 10000000000.0/7, 100000000000000000.0/7, 0.001/7, 0.0001/7, 100000000000000000.0, 0.00000000001,
1.23e-2, 1.234e-5, 1.2345E-10, 1.23456E-20, 5E-20, 1.23E+2, 1.234e5, 1.2345E10, -7.576E-05, 1.23456e20, 5e+20, 9.1093822E-31, 5.9736e24, double.Epsilon };
foreach (int sign in new [] { 1, -1 }) {
foreach (double val in values) {
double val2 = sign * val;
String s1 = val2.ToString("r");
String s2 = ToRoundTrip(val2);
double val2_ = double.Parse(s2);
double diff = Math.Abs(val2 - val2_);
if (diff != 0) {
throw new Exception("Value {0} did not pass ToRoundTrip.".Format2(val.ToString("r")));
}
sb33.AppendLine(s1);
sb33.AppendLine(s2);
sb33.AppendLine();
}
}
}
}
The obligatory Logarithm-based solution. Note that this solution, because it involves doing math, may reduce the accuracy of your number a little bit. Not heavily tested.
private static string DoubleToLongString(double x)
{
int shift = (int)Math.Log10(x);
if (Math.Abs(shift) <= 2)
{
return x.ToString();
}
if (shift < 0)
{
double y = x * Math.Pow(10, -shift);
return "0.".PadRight(-shift + 2, '0') + y.ToString().Substring(2);
}
else
{
double y = x * Math.Pow(10, 2 - shift);
return y + "".PadRight(shift - 2, '0');
}
}
Edit: If the decimal point crosses non-zero part of the number, this algorithm will fail miserably. I tried for simple and went too far.
In the old days when we had to write our own formatters, we'd isolate the mantissa and exponent and format them separately.
In this article by Jon Skeet (https://csharpindepth.com/articles/FloatingPoint) he provides a link to his DoubleConverter.cs routine that should do exactly what you want. Skeet also refers to this at extracting mantissa and exponent from double in c#.
I have just improvised on the code above to make it work for negative exponential values.
using System;
using System.Text.RegularExpressions;
using System.IO;
using System.Text;
using System.Threading;
namespace ConvertNumbersInScientificNotationToPlainNumbers
{
class Program
{
private static string ToLongString(double input)
{
string str = input.ToString(System.Globalization.CultureInfo.InvariantCulture);
// if string representation was collapsed from scientific notation, just return it:
if (!str.Contains("E")) return str;
var positive = true;
if (input < 0)
{
positive = false;
}
string sep = Thread.CurrentThread.CurrentCulture.NumberFormat.NumberDecimalSeparator;
char decSeparator = sep.ToCharArray()[0];
string[] exponentParts = str.Split('E');
string[] decimalParts = exponentParts[0].Split(decSeparator);
// fix missing decimal point:
if (decimalParts.Length == 1) decimalParts = new string[] { exponentParts[0], "0" };
int exponentValue = int.Parse(exponentParts[1]);
string newNumber = decimalParts[0].Replace("-", "").
Replace("+", "") + decimalParts[1];
string result;
if (exponentValue > 0)
{
if (positive)
result =
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
else
result = "-" +
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
}
else // negative exponent
{
if (positive)
result =
"0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Replace("-", "").
Replace("+", "").Length) + newNumber;
else
result =
"-0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Replace("-", "").
Replace("+", "").Length) + newNumber;
result = result.TrimEnd('0');
}
float temp = 0.00F;
if (float.TryParse(result, out temp))
{
return result;
}
throw new Exception();
}
private static string GetZeros(int zeroCount)
{
if (zeroCount < 0)
zeroCount = Math.Abs(zeroCount);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < zeroCount; i++) sb.Append("0");
return sb.ToString();
}
public static void Main(string[] args)
{
//Get Input Directory.
Console.WriteLine(#"Enter the Input Directory");
var readLine = Console.ReadLine();
if (readLine == null)
{
Console.WriteLine(#"Enter the input path properly.");
return;
}
var pathToInputDirectory = readLine.Trim();
//Get Output Directory.
Console.WriteLine(#"Enter the Output Directory");
readLine = Console.ReadLine();
if (readLine == null)
{
Console.WriteLine(#"Enter the output path properly.");
return;
}
var pathToOutputDirectory = readLine.Trim();
//Get Delimiter.
Console.WriteLine("Enter the delimiter;");
var columnDelimiter = (char)Console.Read();
//Loop over all files in the directory.
foreach (var inputFileName in Directory.GetFiles(pathToInputDirectory))
{
var outputFileWithouthNumbersInScientificNotation = string.Empty;
Console.WriteLine("Started operation on File : " + inputFileName);
if (File.Exists(inputFileName))
{
// Read the file
using (var file = new StreamReader(inputFileName))
{
string line;
while ((line = file.ReadLine()) != null)
{
String[] columns = line.Split(columnDelimiter);
var duplicateLine = string.Empty;
int lengthOfColumns = columns.Length;
int counter = 1;
foreach (var column in columns)
{
var columnDuplicate = column;
try
{
if (Regex.IsMatch(columnDuplicate.Trim(),
#"^[+-]?[0-9]+(\.[0-9]+)?[E]([+-]?[0-9]+)$",
RegexOptions.IgnoreCase))
{
Console.WriteLine("Regular expression matched for this :" + column);
columnDuplicate = ToLongString(Double.Parse
(column,
System.Globalization.NumberStyles.Float));
Console.WriteLine("Converted this no in scientific notation " +
"" + column + " to this number " +
columnDuplicate);
}
}
catch (Exception)
{
}
duplicateLine = duplicateLine + columnDuplicate;
if (counter != lengthOfColumns)
{
duplicateLine = duplicateLine + columnDelimiter.ToString();
}
counter++;
}
duplicateLine = duplicateLine + Environment.NewLine;
outputFileWithouthNumbersInScientificNotation = outputFileWithouthNumbersInScientificNotation + duplicateLine;
}
file.Close();
}
var outputFilePathWithoutNumbersInScientificNotation
= Path.Combine(pathToOutputDirectory, Path.GetFileName(inputFileName));
//Create Directory If it does not exist.
if (!Directory.Exists(pathToOutputDirectory))
Directory.CreateDirectory(pathToOutputDirectory);
using (var outputFile =
new StreamWriter(outputFilePathWithoutNumbersInScientificNotation))
{
outputFile.Write(outputFileWithouthNumbersInScientificNotation);
outputFile.Close();
}
Console.WriteLine("The transformed file is here :" +
outputFilePathWithoutNumbersInScientificNotation);
}
}
}
}
}
This code takes an input directory and based on the delimiter converts all values in scientific notation to numeric format.
Thanks
try this one:
public static string DoubleToFullString(double value,
NumberFormatInfo formatInfo)
{
string[] valueExpSplit;
string result, decimalSeparator;
int indexOfDecimalSeparator, exp;
valueExpSplit = value.ToString("r", formatInfo)
.ToUpper()
.Split(new char[] { 'E' });
if (valueExpSplit.Length > 1)
{
result = valueExpSplit[0];
exp = int.Parse(valueExpSplit[1]);
decimalSeparator = formatInfo.NumberDecimalSeparator;
if ((indexOfDecimalSeparator
= valueExpSplit[0].IndexOf(decimalSeparator)) > -1)
{
exp -= (result.Length - indexOfDecimalSeparator - 1);
result = result.Replace(decimalSeparator, "");
}
if (exp >= 0) result += new string('0', Math.Abs(exp));
else
{
exp = Math.Abs(exp);
if (exp >= result.Length)
{
result = "0." + new string('0', exp - result.Length)
+ result;
}
else
{
result = result.Insert(result.Length - exp, decimalSeparator);
}
}
}
else result = valueExpSplit[0];
return result;
}
Being millions of programmers world wide, it's always a good practice to try search if someone has bumped into your problem already. Sometimes there's solutions are garbage, which means it's time to write your own, and sometimes there are great, such as the following:
http://www.yoda.arachsys.com/csharp/DoubleConverter.cs
(details: http://www.yoda.arachsys.com/csharp/floatingpoint.html)
string strdScaleFactor = dScaleFactor.ToString(); // where dScaleFactor = 3.531467E-05
decimal decimalScaleFactor = Decimal.Parse(strdScaleFactor, System.Globalization.NumberStyles.Float);
I don't know if my answer to the question can still be helpful. But in this case I suggest the "decomposition of the double variable into decimal places" to store it in an Array / Array of data of type String.
This process of decomposition and storage in parts (number by number) from double to string, would basically work with the use of two loops and an "alternative" (if you thought of workaround, I think you got it), where the first loop will extract the values from double without converting to String, resulting in blessed scientific notation and storing number by number in an Array. And this will be done using MOD - the same method to check a palindrome number, which would be for example:
String[] Array_ = new double[ **here you will put an extreme value of places your DOUBLE can reach, you must have a prediction**];
for (int i = 0, variableDoubleMonstrous > 0, i++){
x = variableDoubleMonstrous %10;
Array_[i] = x;
variableDoubleMonstrous /= 10;
}
And the second loop to invert the Array values (because in this process of checking a palindrome, the values invert from the last place, to the first, from the penultimate to the second and so on. Remember?) to get the original value:
String[] ArrayFinal = new String[the same number of "places" / indices of the other Array / Data array];
int lengthArray = Array_.Length;
for (int i = 0, i < Array_.Length, i++){
FinalArray[i] = Array_[lengthArray - 1];
lengthArray--;
}
***Warning: There's a catch that I didn't pay attention to. In that case there will be no "." (floating point decimal separator or double), so this solution is not generalized. But if it is really important to use decimal separators, unfortunately the only possibility (If done well, it will have a great performance) is:
**Use a routine to get the position of the decimal point of the original value, the one with scientific notation - the important thing is that you know that this floating point is before a number such as the "Length" position x, and after a number such as the y position - extracting each digit using the loops - as shown above - and at the end "export" the data from the last Array to another one, including the decimal place divider (the comma, or the period , if variable decimal, double or float) in the imaginary position that was in the original variable, in the "real" position of that matrix.
*** The concept of position is, find out how many numbers occur before the decimal point, so with this information you will be able to store in the String Array the point in the real position.
NEEDS THAT CAN BE MADE:
But then you ask:
But what about when I'm going to convert String to a floating point value?
My answer is that you use the second matrix of this entire process (the one that receives the inversion of the first matrix that obtains the numbers by the palindrome method) and use it for the conversion, but always making sure, when necessary, of the position of the decimal place in future situations, in case this conversion (Double -> String) is needed again.
But what if the problem is to use the value of the converted Double (Array of Strings) in a calculation. Then in this case you went around in circles. Well, the original variable will work anyway even with scientific notation. The only difference between floating point and decimal variable types is in the rounding of values, which depending on the purpose, it will only be necessary to change the type of data used, but it is dangerous to have a significant loss of information, look here
I could be wrong, but isn't it like this?
data.ToString("n");
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
i think you need only to use IFormat with
ToString(doubleVar, System.Globalization.NumberStyles.Number)
example:
double d = double.MaxValue;
string s = d.ToString(d, System.Globalization.NumberStyles.Number);
My solution was using the custom formats.
try this:
double d;
d = 1234.12341234;
d.ToString("#########0.#########");
Just to build on what jcasso said what you can do is to adjust your double value by changing the exponent so that your favorite format would do it for you, apply the format, and than pad the result with zeros to compensate for the adjustment.
This works fine for me...
double number = 1.5E+200;
string s = number.ToString("#");
//Output: "150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string num = db.SelectNums(id);
string[] numArr = num.Split('-').ToArray();
string num contain for an example "48030-48039";
string[] numArr will therefor contain (48030, 48039).
Now I have two elements, a high and low. I now need to get ALL the numbers from 48030 to 48039. The issue is that it has to be string since there will be telephone numbers with leading zeroes now and then.
Thats why I cannot use Enumerable.Range().ToArray() for this.
Any suggestions? The expected result should be 48030, 48031, 48032, ... , 48039
This should work with your leading zero requirement:
string num = db.SelectNums(id);
string[] split = num.Split('-');
long start = long.Parse(split[0]);
long end = long.Parse(split[1]);
bool includeLeadingZero = split[0].StartsWith("0");
List<string> results = new List<string>();
for(int i = start; i <= end; i++)
{
string result = includeLeadingZero ? "0" : "";
result += i.ToString();
results.Add(result);
}
string[] arrayResults = results.ToArray();
A few things to note:
It assumes your input will be 2 valid integers split by a single hyphen
I am giving you a string array result, personally I would prefer to work with a List<int> in the end
If the first number contains a single leading zero, then all results will contain a single leading zero
It uses long to cater for longer numbers, beware that the max number that will parse is 9,223,372,036,854,775,807, you could double this value (not length) by using ulong
Are you saying this?
int[] nums = new int[numArr.Length];
for (int i = 0; i < numArr.Length; i++)
{
nums[i] = Convert.ToInt32(numArr[i]);
}
Array.Sort(nums);
for (int n = nums[0]; n <= nums[nums.Length - 1]; n++)
{
Console.WriteLine(n);
}
here link
I am expecting your string always have valid two integers, so using Parse instead TryParse
string[] strList = "48030-48039".Split('-').ToArray();
var lst = strList.Select(int.Parse).ToList();
var min = lst.OrderBy(l => l).FirstOrDefault();
var max = lst.OrderByDescending(l => l).FirstOrDefault();
var diff = max - min;
//adding 1 here otherwise 48039 will not be there
var rng = Enumerable.Range(min,diff+1);
If you expecting invalid string
var num = 0;
var lst = (from s in strList where int.TryParse(s, out num) select num).ToList();
This is one way to do it:
public static string[] RangeTest()
{
Boolean leadingZero = false;
string num = "048030-48039"; //db.SelectNums(id);
if (num.StartsWith("0"))
leadingZero = true;
int min = int.Parse(num.Split('-').Min());
int count = int.Parse(num.Split('-').Max()) - min;
if (leadingZero)
return Enumerable.Range(min, count).Select(x => "0" + x.ToString()).ToArray();
else
return Enumerable.Range(min, count).Select(x => "" + x.ToString()).ToArray(); ;
}
You can use string.Format to ensure numbers are formatted with leading zeros. That will make the method work with arbitrary number of leading zeros.
private static string CreateRange(string num)
{
var tokens = num.Split('-').Select(s => s.Trim()).ToArray();
// use UInt64 to allow huge numbers
var start = UInt64.Parse(tokens[0]);
var end = UInt64.Parse(tokens[1]);
// if your start number is '000123', this will create
// a format string with 6 zeros ('000000')
var format = new string('0', tokens[0].Length);
// use StringBuilder to make GC happy.
// (if only there was a Enumerable.Range<ulong> overload...)
var sb = new StringBuilder();
for (var i = start; i <= end; i++)
{
// format ensures that your numbers are padded properly
sb.Append(i.ToString(format));
sb.Append(", ");
}
// trim trailing comma after the last element
if (sb.Length >= 2) sb.Length -= 2;
return sb.ToString();
}
Usage example:
// prints 0000012, 0000013, 0000014
Console.WriteLine( CreateRange("0000012-0000014") );
Three significant issues were brought up in comments:
The phone numbers have enough digits to exceed Int32.MaxValue so
converting to int isn't viable.
The phone numbers can have leading zeros (presumeably for some
international calling?)
The possible range of numbers can theoretically exceed the maximum size of an array (which may have memory issues, and I think may not be represented as a string)
As such, you may need to use long instead of int, and I would suggest using deferred execution if needed for very large ranges.
public static IEnumerable<string> EnumeratePhoneNumbers(string fromAndTo)
{
var split = fromAndTo.Split('-');
return EnumeratePhoneNumbers(split[0], split[1]);
}
public static IEnumerable<string> EnumeratePhoneNumbers(string fromString, string toString)
{
long from = long.Parse(fromString);
long to = long.Parse(toString);
int totalNumberLength = fromString.Length;
for (long phoneNumber = from; phoneNumber <= to; phoneNumber++)
{
yield return phoneNumber.ToString().PadLeft(totalNumberLength, '0');
}
}
This assumes that the padded zeros are already included in the lower bound fromString text. It will iterate and yield out numbers as you need them. This can be useful if you're churning out a lot of numbers and don't need to fill up memory with them, or if you just need the first 10 or 100. For example:
var first100Numbers =
EnumeratePhoneNumbers("0018155500-7018155510")
.Take(100)
.ToArray();
Normally that range would produce 7 billion results which cannot be stored in an array, and might run into memory issues (I'm not even sure if it can be stored in a string); by using deferred execution, you only create the 100 needed.
If you do have a small range, you can still join up your results into a string as you desired:
string numberRanges = String.Join(", ", EnumeratePhoneNumbers("0018155500-0018155510"));
And naturally you can put this array creation into your own helper method:
public static string GetPhoneNumbersListing(string fromAndTo)
{
return String.Join(", ", EnumeratePhoneNumbers("0018155500-0018155510"));
}
So your usage would be:
string numberRanges = GetPhoneNumbersListing("0018155500-0018155510");
A complete solution inspired by the answer by #Dan-o:
Inputs:
Start: 48030
End: 48039
Digits: 6
Expected String Output:
048030, 048031, 048032, 048033, 048034, 048035, 048036, 048037, 048038, 048039
Program:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
public class Program
{
public static void Main()
{
int first = 48030;
int last = 48039;
int digits = 6;
Console.WriteLine(CreateRange(first, last, digits));
}
public static string CreateRange(int first, int last, int numDigits)
{
string separator = ", ";
var sb = new StringBuilder();
sb.Append(first.ToString().PadLeft(numDigits, '0'));
foreach (int num in Enumerable.Range(first + 1, last - first))
{
sb.Append(separator + num.ToString().PadLeft(numDigits, '0'));
}
return sb.ToString();
}
}
For Each item In Enumerable.Range(min, count).ToArray()
something = item.PadLeft(5, "0")
Next
I am looking for a method that adds zero's up to 16 characters before a decimal, and a minus sign if the value is minus. E.g.,
18,52 becomes 000000000000001852, and
-18,52 becomes-00000000000001852
I have an idea how to implement this by using replacements and if-statements, and using the PadLeft method where characters are padded to the left to what length you specify. But I am not sure how to make it exactly.
What I have right now is this:
static string FormatDecimal(decimal d, int length = 0, char a = '0')
{
var sb = new StringBuilder();
var rounded = decimal.Round(d, 2, MidpointRounding.AwayFromZero);
if (rounded < 0)
{
sb.Append("-");
}
else
{
sb.Append("");
}
var lastPart = rounded.ToString().Replace(",", "").Replace(".", "");
var lengthMiddle = length - sb.Length - lastPart.Length;
for (int i = 0; i < lengthMiddle; i++)
{
sb.Append(a);
}
sb.Append(lastPart);
return sb.ToString();
}
When I look at the code and do Console.WriteLine(FormatDecimal(-18m, 16, '0')) I see that
The code is 1. very long, and 2. it does not work... The rounding fails and just keeps the -18 and no minus sign is added.
I would be very grateful if someone could help me out with this one!
If you want to represent two decimal digits in your string and eliminate the decimal mark, you can simply multiply your number by 100. To pad up to 16 digits, use the D format string:
decimal d = -18.52m;
string s = ((int)(d * 100)).ToString("D16");
Edit: If you only want to pad up to 15 digits for negative numbers, you could use a conditional:
decimal d = -18.52m;
int i = (int)(d * 100);
string s = i.ToString(i >= 0 ? "D16" : "D15");
Alternatively, you could express the conditional within the format string itself using section separators:
string s = i.ToString("0000000000000000;-000000000000000");
Instead of building this yourself, use what's already there. decimal.ToString() will format a number for you:
decimal d = 18.52;
string s = d.ToString("0000000000.##"); // = "0000000018.52"
decimal d = 18.00;
string s = d.ToString("0000000000.##"); // = "0000000018"
You could build your format string using the string constructor:
int length = 10;
string formatString = string.Concat(new string('0', length), ".##")
string s = d.ToString(formatString); // = "0000000018"
Note: this doesn't take care of your rounding, but I'm not clear from the question what your requirement is there.
I need to format a double value so that it fits within a field of 13 characters. Is there a way to do this with String.Format or am I stuck with character-by-character work?
Edits: (hopefully they will stay this time)
With cases greater than a trillion I am to report an error. It's basically a calculator interface.
My own answer:
private void DisplayValue(double a_value)
{
String displayText = String.Format("{0:0." + "".PadRight(_maxLength, '#') + "}", a_value);
if (displayText.Length > _maxLength)
{
var decimalIndex = displayText.IndexOf('.');
if (decimalIndex >= _maxLength || decimalIndex < 0)
{
Error();
return;
}
var match = Regex.Match(displayText, #"^-?(?<digits>\d*)\.\d*$");
if (!match.Success)
{
Error();
return;
}
var extra = 1;
if (a_value < 0)
extra = 2;
var digitsLength = match.Groups["digits"].Value.Length;
var places = (_maxLength - extra) - digitsLength;
a_value = Math.Round(a_value, places);
displayText = String.Format("{0:0." + "".PadRight(_maxLength, '#') + "}", a_value);
if (displayText.Length > _maxLength)
{
Error();
return;
}
}
DisplayText = displayText;
}
If this is calculator, then you can not use character-by-character method you mention in your question. You must round number to needed decimal places first and only then display it otherwise you could get wrong result. For example, number 1.99999 trimmed to length of 4 would be 1.99, but result 2 would be more correct.
Following code will do what you need:
int maxLength = 3;
double number = 1.96;
string output = null;
int decimalPlaces = maxLength - 2; //because every decimal contains at least "0."
bool isError = true;
while (isError && decimalPlaces >= 0)
{
output = Math.Round(number, decimalPlaces).ToString();
isError = output.Length > maxLength;
decimalPlaces--;
}
if (isError)
{
//handle error
}
else
{
//we got result
Debug.Write(output);
}
You have a lot formatting options using String.Format, just specify format after placeholder like this {0:format}.
Complete example looks like this:
Console.WriteLine("Your account balance is {0:N2}.", value);
Output would be:
Your account balance is 15.34.
All of the options for numeric types are listed here:
http://msdn.microsoft.com/en-us/library/dwhawy9k(v=vs.110).aspx
This seems to work for me (but is hand-rolled):
static string FormatDouble(double d)
{
int maxLen = 13;
double threshold = Math.Pow(10, maxLen);
if (d >= threshold || d <= 0 - (threshold/10))
return "OVERFLOW";
string strDisplay = "" + d;
if (strDisplay.Length > maxLen )
strDisplay = strDisplay.Substring(0, maxLen);
if (strDisplay.EndsWith("."))
strDisplay = strDisplay.Replace(".", "");
return strDisplay;
}
Let me know if it gives you trouble with scientific notation creeping in. I believe the format "{0:R}" should help you avoid that explicitly.
Also, I wasn't sure if you were including +/- sign in digit count or if that was in a separate UI slot.
The theory on rounding here is that, yes, "" + d might round some things, but in general it's going to be many more digits out than are ever displayed so it shouldn't matter. So this method should always truncate.
Here's a solution that does rounding. (I couldn't think of a non-mathematical way to do it):
static string FormatDouble(double d)
{
int maxLen = 13;
int places = (int)Math.Max(Math.Log10(Math.Abs(d)), 0);
places += (d == Math.Abs(d) ? 1 : 2);
if (places > maxLen || places < 1 - maxLen)
return "OVERFLOW";
if (Math.Floor(d) == d) ++places; // no decimal means one extra spot
d = Math.Round(d, Math.Max(maxLen - places - 1, 0));
return string.Format("{0:R}", d);
}
Note: I still think your users might appreciate seeing something closer to what is being stored in the underlying memory than what is often typical of calculators. (I especially hate the ones that can turn 0.99 into 1.01) Either way, you've got at least 3 solutions now so it's up to you.