I am converting a string like "41.00027357629127", and I am using;
Convert.ToSingle("41.00027357629127");
or
float.Parse("41.00027357629127");
These methods return 4.10002732E+15.
When I convert to float I want "41.00027357629127". This string should be the same...
Your thread's locale is set to one in which the decimal mark is "," instead of ".".
Try using this:
float.Parse("41.00027357629127", CultureInfo.InvariantCulture.NumberFormat);
Note, however, that a float cannot hold that many digits of precision. You would have to use double or Decimal to do so.
You can use the following:
float asd = (float) Convert.ToDouble("41.00027357629127");
First, it is just a presentation of the float number you see in the debugger. The real value is approximately exact (as much as it's possible).
Note: Use always CultureInfo information when dealing with floating point numbers versus strings.
float.Parse("41.00027357629127",
System.Globalization.CultureInfo.InvariantCulture);
This is just an example; choose an appropriate culture for your case.
Use Convert.ToDouble("41.00027357629127");
Convert.ToDouble documentation
The precision of float is 7 digits. If you want to keep the whole lot, you need to use the double type that keeps 15-16 digits. Regarding formatting, look at a post about formatting doubles. And you need to worry about decimal separators in C#.
A 2022 way of converting an string that represents a float value:
(float)Convert.ToDecimal(value, CultureInfo.GetCultureInfo("en-US"));
where you also can choose what kind of float are you expecting to convert, because some CultureInfo instances represents decimal values with a , and others with ..
If you need more decimals to obtain more precision, just not use float
Convert.ToDecimal(value, CultureInfo.GetCultureInfo("en-US"));
You can double.Parse("41.00027357629127");
First you need to using System.Globalization to dealing convertions from string to float/double/decimal without problem.
Then you can call Parse on float(or double/decimal depending at the accuracy you need), and as argument in Parse you need your string (you can store it in a variable if you want) and CultureInfo.InvariantCulture.NumberFormat
So, as previous users already explained:
float.Parse("41.00027357629127", CultureInfo.InvariantCulture.NumberFormat);
You can use parsing with double instead of float to get more precision value.
Related
I am trying to round decimal number upto two decimal places which is working perfectly.
I am doing as below :
Math.Round(Amount, 2)
So, if I have Amount as 40000.4567, I am getting 40000.46which is exactly what I want.
Now problem is I have decimal number like 40000.0000, when I round it, the result is 40000, and what I really want is 40000.00. So round will always neglect trailing zeros.
To solve this problem, I have the option of converting it to string and use format , but I don't want to do that as that will be inefficient and I believe there must be some way to do it better.
I also tried something like
Decimal.Round(Amount, 2)
Now one way can be to check whether number contains anything in fractional part and use round function accordingly , but that is really bad way to do it.
I can't use truncate as well due to obvious reasons of this being related to amount.
What is the way around?
It is rounding correctly but you fail to understand that the value is not the format. There is no difference between the two values, 40000 and 40000.00, and you'll have a similar issue with something like 3.1.
Simply use formatting to output whatever number you have to two decimal places, such as with:
Console.WriteLine(String.Format("{0:0.00}", value));
or:
Console.WriteLine(value.ToString("0.00"));
You are mixing two things - rounding and output formatting. In order to output a number in a format you want you can use function string.Format with required format, for example:
decimal number = 1234.567m;
string.Format("{0:#.00}", number);
You can read more about custom numeric format strings in MSDN
I think what you're looking for is displaying two decimals, even if they are zero. You can use string.Format for this (I've also combined it with Round):
Console.WriteLine(string.Format("{0:0.00}", Math.Round(Amount, 2));
for rounding decimal number you can use
decimal number=200.5555m;
number= Math.Round(number, 2);
string numString= string.Format("{0:0.00}", number);
I want to reduce the float type precision from 7 digits to 6 after the "." I tried multiplying the number by 10 but this didn't work. Any ideas?
If you're only trying to format the number on output (ie. in conversion to a string), you just need to use a proper format string:
13.651234f.ToString("f6"); // Always six decimal places
If you need to do that for your application logic, you probably want to use decimal rather than float - float is a binary number, so the notion of "decimal" decimal places is a bit off.
I have some float values I want to convert to a string, I want to keep the formatting the same when converting, i.e. 999.0000(float) -> 999.0000(String). My problem is when the values contain an arbitrary number of zeroes after the decimal point, as in the previous example, they are stripped away when converting to a string, so the result I actually end up with is 999.
I looked at the format specifiers for the toString() method on MSDN, the RoundTrip ('R') specifier looks like it will produce what I want, but it is only supported for Single, Double and BigInt variables. Is there a format specifier like this for float variables?? Or would it be easier to just convert the values to doubles?
UPDATE: Just for clarity, the reason why I want to keep the trailing zeroes is because I'm doing a comparison of decimal places, i.e. I'm comparing the number of digits after the decimal place between two values. So for example, 1.00 and 1.00000 have a different number of digits after the decimal point. I know it's a strange request, it's for work and the requirement is coming from on high.
UPDATE 2-3-11:
I was thinking about this too hard, I'm reading the numbers from a txt file and then parsing them as floats, I'm going to modify the program to check whether the string values are decimals or whole numbers. Sorry for wasting your time, although this was very insightful.
Use ToString() with this format:
12345.678901.ToString("0.0000"); // outputs 12345.6789
12345.0.ToString("0.0000"); // outputs 12345.0000
Put as much zero as necessary at the end of the format.
Firstly, as Etienne says, float in C# is Single. It is just the C# keyword for that data type.
So you can definitely do this:
float f = 13.5f;
string s = f.ToString("R");
Secondly, you have referred a couple of times to the number's "format"; numbers don't have formats, they only have values. Strings have formats. Which makes me wonder: what is this thing you have that has a format but is not a string? The closest thing I can think of would be decimal, which does maintain its own precision; however, calling simply decimal.ToString should have the effect you want in that case.
How about including some example code so we can see exactly what you're doing, and why it isn't achieving what you want?
You can pass a format string to the ToString method, like so:
ToString("N4"); // 4 decimal points Number
If you want to see more modifiers, take a look at MSDN - Standard Numeric Format Strings
In C#, float is an alias for System.Single (a bit like intis an alias for System.Int32).
So, WPF calls ToString() on objects when generating TextColumns in DataGrid and then i found out strange thing about ToString() method:
Check this out :
object a = 0.3780000001;//Something like this
Console.WriteLine(a.ToString());//Gets truncated in some cases
First, I thought it was just rounding, but few times I was able to reproduce such behavior on
doubles with < 15 digits after dot. Am I missing something?
To the computer, 0.378 and 0.378000...0001 are the same number. See this question: Why is floating point arithmetic in C# imprecise?
As defined on the MSDN page for System.Double, the double type only contains a maximum of fifteen digits of precision. Even though it maintains 17 internally, your figure contains 18 significant digits; this is outside the range of System.Double.
Use decimal instead of float for a more precise type.
By default, Double.ToString() truncates to 15 digits after the dot, but if you really want to use the double data type and you need those 2 extra digits, you can use th "G17" formatting string:
double x = 3.1415926535897932;
string pi = x.ToString("G17");
This will give you a string with the full 17 digits.
I wouldn't assume (so fast) that you found a bug in something as crucial as C#'s ToString implementation.
The behaviour you're experiencing is caused by the fact that a float is imprecisely stored in computer memory (also see this question).
maybe the number format's accuracy range doesn't contain that number? (ie, float only has accuracy to a few significant figures)
If you're data-binding the value, you can supply a ValueConverter which formats the number any way you want.
http://msdn.microsoft.com/en-us/library/system.windows.data.ivalueconverter.aspx
Set a to be an Decimal and it will print it correctly!
decimal a = 0.378000000000000001m;
Console.WriteLine(a.ToString());
You could have a common decimal format setting to use all the time.
eg
object a = 0.378000000000000001;
Console.WriteLine(a.ToString(Settings.DecimalFormat));
In c# double type how can i set the number of digits after the point, i need only 4.
thank you.
You can't. Binary floating point doesn't work like that. You can format a double that way (e.g. using "f4" as the format string), but if you're dealing with values which have a natural number of decimal places, then you should probably be using decimal instead. Could you give us more information about what your values represent?
You can't set the number of digits after the point on the double directly.
You can change the string representation of the double using a format string.
One example would be:
string.Format("{0:0.####}", number);
Or as Jon Skeet points out:
number.ToString("f4")
Use this to compare two floating point numbers to 4 digits in the fraction:
if (Math.Abs(a - b) < 1E-4) {
// close enough
//...
}