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"
I have strings that look like this:
1.23.4.34
12.4.67
127.3.2.21.3
1.1.1.9
This is supposed to be a collection of numbers, separated by '.' symbols, similar to an ip address. I need to increment only the last digit/digits.
Expected Output:
1.23.4.35
12.4.68
127.3.2.21.4
1.1.1.10
Basically, increment whatever the number that is after the last '.' symbol.
I tried this:
char last = numberString[numberString.Length - 1];
int number = Convert.ToInt32(last);
number = number + 1;
If I go with the above code, I just need to replace the characters after the last '.' symbol with the new number. How do I get this done, good folks? :)
It seems to me that one method would be to:
split the string on . to get an array of components.
turn the final component into an integer.
increment that integer.
turn it back into a string.
recombine the components with . characters.
See, for example, the following program:
using System;
namespace ConsoleApplication1 {
class Program {
static void Main(string[] args) {
String original = "1.23.4.34";
String[] components = original.Split('.');
int value = Int32.Parse(components[components.Length - 1]) + 1;
components[components.Length - 1] = value.ToString();
String newstring = String.Join(".",components);
Console.WriteLine(newstring);
}
}
}
which outputs the "next highest" value of:
1.23.4.35
You can use string.LastIndexOf().
string input = "127.3.2.21.4";
int lastIndex = input.LastIndexOf('.');
string lastNumber = input.Substring(lastIndex + 1);
string increment = (int.Parse(lastNumber) + 1).ToString();
string result = string.Concat(input.Substring(0, lastIndex + 1), increment);
You need to extract more than just the last character. What if the last character is a 9 and then you add 1 to it? Then you need to correctly add one to the preceding character as well. For example, the string 5.29 should be processed to become 5.30 and not simply 5.210 or 5.20.
So I suggest you split the string into its number sections. Parse the last section into an integer. Increment it and then create the string again. I leave it as an exercise for the poster to actually write the few lines of code. Good practice!
Something like this:
var ip = "1.23.4.34";
var last = int.Parse(ip.Split(".".ToCharArray(),
StringSplitOptions.RemoveEmptyEntries).Last());
last = last + 1;
ip = string.Format("{0}.{1}",ip.Remove(ip.LastIndexOf(".")) , last);
If you are dealing with IP, there will be some extra code in case of .034, which should be 035 instead of 35. But that logic is not that complicated.
It's simple as this, use Split() and Join() String methods
String test = "1.23.4.34"; // test string
String[] splits = test.Split('.'); // split by .
splits[splits.Length - 1] = (int.Parse(splits[splits.Length - 1])+1).ToString(); // Increment last integer (Note : Assume all are integers)
String answ = String.Join(".",splits); // Use string join to make the string from string array. uses . separator
Console.WriteLine(answ); // Answer : 1.23.4.35
Using a bit of Linq
int[] int_arr = numberString.Split('.').Select(num => Convert.ToInt32(num)).ToArray();
int_arr[int_arr.Length - 1]++;
numberString = "";
for(int i = 0; i < int_arr.Length; i++) {
if( i == int_arr.Length - 1) {
numberString += int_arr[i].ToString();
}
else {
numberString += (int_arr[i].ToString() + ".");
}
}
Note: on phone so can't test.
My Solution is:
private static string calcNextCode(string value, int index)
{
if (value is null) return "1";
if (value.Length == index + 1) return value + "1";
int lastNum;
int myIndex = value.Length - ++index;
char myValue = value[myIndex];
if (int.TryParse(myValue.ToString(), NumberStyles.Integer, null, out lastNum))
{
var aStringBuilder = new StringBuilder(value);
if (lastNum == 9)
{
lastNum = 0;
aStringBuilder.Remove(myIndex, 1);
aStringBuilder.Insert(myIndex, lastNum);
return calcNextCode(aStringBuilder.ToString(), index++);
}
else
{
lastNum++;
}
aStringBuilder.Remove(myIndex, 1);
aStringBuilder.Insert(myIndex, lastNum);
return aStringBuilder.ToString();
}
return calcNextCode(value, index++);
}
It tells me that it can't convert int to bool.
Tried TryParse but for some reason the argument list is invalid.
Code:
private void SetNumber(string n)
{
// if user input is a number then
if (int.Parse(n))
{
// if user input is negative
if (h < 0)
{
// assign absolute version of user input
number = Math.Abs(n);
}
else
{
// else assign user input
number = n;
}
}
else
{
number = 0; // if user input is not an int then set number to 0
}
}
You were probably very close using TryParse, but I'm guessing you forgot the out keyword on the parameter:
int value;
if (int.TryParse(n, out value))
{
}
Just use this:
int i;
bool success = int.TryParse(n, out i);
if the parse was successful, success is true.
If that case i contain the number.
You probably got the out argument modifier wrong before. It has the out modifier to indicate that it is a value that gets initialized within the method called.
You can try with some simple regular expression :
bool IsNumber(string text)
{
Regex regex = new Regex(#"^[-+]?[0-9]*\.?[0-9]+$");
return regex.IsMatch(text);
}
private void SetNumber(string n)
{
int nVal = 0;
if (int.TryParse(n, out nVal))
{
if (nVal < 0)
number = Math.Abs(nVal);
else
number = nVal;
}
else
number = 0;
}
There are a lot of problems with this code:
Using VB-style line comments (') instead of C# slashes
Parse for integer returns an int and not a bool
You should use TryParse with an out value
h does not seem to be valid at all. Is it a type for n?
There are variables that do not seem to be defined in function scope (number) are they defined at class scope?
But try this:
private void SetNumber(string n)
{
int myInt;
if (int.TryParse(n, out myInt)) //if user input is a number then
{
if (myInt < 0) //if user input is negative
number = Math.Abs(n); //assign absolute version of user input
else //else assign user input
number = n;
}
else number = 0; //if user input is not an int then set number to 0
}
You could try something like below using int.TryParse..
private void SetNumber(string n)
{
int parsed = -1;
if (int.TryParse(n, out parsed)) //if user input is a number then
...
The reason there are complaints that it cannot convert an int to a bool is because the return type of int.Parse() is an int and not a bool and in c# conditionals need to evaluate bool values.
int.Parse will give you back an integer rather than a boolean.
You could use int.TryParse as you suggested.
int parsedValue;
if(int.TryParse(n, out parsedValue))
{
}
Well for one thing the inner if statement has an 'h' instead of an 'n' if(h < 0). But TryParse should work there assuming that 'number' is a class variable.
private void SetNumber(string n)
{
int temp;
bool success = Int32.TryParse(n, out temp);
// If conversion successful
if (success)
{
// If user input is negative
if (temp < 0)
number = Math.Abs(temp); // Assign absolute version of user input
else // Assign user input
number = temp;
}
else
{
number = 0;
}
}
int.Parse will convert a string to an integer. Current you have it within an if statement, so its treating the returned value of int.Parse as a bool, which its not.
Something like this will work:
private void SetNumber(string n)
{
int num = 0;
try{
num = int.Parse(n);
number = Math.Abs(num);
}catch(Exception e){
number = 0;
}
}
I did this in the simplest way I knew how.
static void Main(string[] args)
{
string a, b;
int f1, f2, x, y;
Console.WriteLine("Enter two inputs");
a = Convert.ToString(Console.ReadLine());
b = Console.ReadLine();
f1 = find(a);
f2 = find(b);
if (f1 == 0 && f2 == 0)
{
x = Convert.ToInt32(a);
y = Convert.ToInt32(b);
Console.WriteLine("Two inputs r number \n so tha additon of these text box is= " + (x + y).ToString());
}
else
Console.WriteLine("One or tho inputs r string \n so tha concadination of these text box is = " + (a + b));
Console.ReadKey();
}
static int find(string s)
{
string s1 = "";
int f;
for (int i = 0; i < s.Length; i++)
for (int j = 0; j <= 9; j++)
{
string c = j.ToString();
if (c[0] == s[i])
{
s1 += c[0];
}
}
if (s==s1)
f= 0;
else
f= 1;
return f;
}