Display double as its true value? - c#

I have a double in C# like below:
var myDouble = 1.0;
When I want to display this double, it prints as 1 when it should print as 1.0, I'm using this for a version so it kind of defeats the object.
How can I display it as its true value? 1.0 not 1...

Use
var myDouble = 1.0;
Console.WriteLine(myDouble.ToString("0.0"));

This question does not technically have an answer. The double is a 64-bit (and float as 32-bit) numeric floating point type. There's fundamentally no distinction between 1, 1.0, 1.00 and 1.000. These are all the SAME value.
In fact, that binary value if its 64-bit comes out to 3FF0000000000000 for ALL those examples. If its 32-bit, the binary expression is different but the idea is the same.
The format is defined by IEE 754 specification. More information on that is here:
https://en.wikipedia.org/wiki/IEEE_754
There are 3 sections if a floating point number:
1. Sign Bit
2. Exponent
3. Mantissa
There's no section for "number of zeros I typed".
Basically though, once you convert it to a number, you have lost the definition of how you had formatted the text, that just doesn't exist for a number.
If you want to remember how many decimal places you typed, you need to store it as a string. If you just want to format the decimal data type, you need to specify a format as there's no knowledge of your original format.

It is just as correct to say that "1" is its true value.
You can format the number as you want using something like this:
myDouble.ToString("0.0");
But I really don't understand why you would. A version is generally formatted a specific way. Some people format it as "1.00". I really think it makes more sense to store it in a string so it is always guaranteed to appear exactly as it should.

You should declare the variable in double type like this:
double a = 1.0;
Console.WriteLine(string.Format("{0:0.0}",a));

Related

set precision c#

Is it possible to set precision of double values in c# for all double values included in the project? I have a lot of values there and changing them with Math.Round would be exhausting. I need to have the double value as 5.12345 instead of 5.123455123321321 for example.
This is a fundamental limitation of floating point types.. double is actually store internally as a sign exponent and mantissa, the exponent is base 2, so has a lot of trouble dealing with base 10..
The easiest solution is to use a base 10 64bit floating point type, namely decimal. Its still floating point, it still only limited precision but it is a lot friendly and more accurate to work with in a lot of cases
Update
If all you want to do is change the display output, you can either use rounding (which you know), or the appropriate format specifiers with string.format ToString or string interpolation
Example
var number = 5.123455123321321;
Console.WriteLine(number.ToString("F3",
CultureInfo.InvariantCulture));
Console.WriteLine($"{number:F3}");
// Displays 5.123
// Displays 5.123

Double to String Format text format

i have the follwing lines of code
double formId=2013519115027601;
txtEditFormID.Text = formid.ToString();
it gives me output
2.0135191150276E+15
if i write
txtEditFormID.Text = formId.ToString("0.0", CultureInfo.InvariantCulture);
it gives me
2013519115027600.0
but i want the label text
2013519115027601
how to do it?
I don't have enough information about the usage of your formId variable.
As it is shown above it seems an error to use a double datatype when there is no decimals to work on. So redefining your variable as a long datatype will be easy and the conversion will be the same.
long formId=2013519115027601;
txtEditFormID.Text = formid.ToString();
Not to mention the added benefit to your code to work with whole numbers instead of floating point numbers.
However, if you want to maintain the current datatype then
txtEditFormID.Text = formId.ToString("R");
The Round Trip Format Specifier
When a Single or Double value is formatted using this specifier, it is
first tested using the general format, with 15 digits of precision for
a Double and 7 digits of precision for a Single. If the value is
successfully parsed back to the same numeric value, it is formatted
using the general format specifier. If the value is not successfully
parsed back to the same numeric value, it is formatted using 17 digits
of precision for a Double and 9 digits of precision for a Single.
Your first option is to use data type as long or decimal . Something else you can do if you want to keep using double is this :
double formId = 2013519115027601;
string text = formId.ToString();
txtEditFormID.Text = text.Replace(".",string.Empty);
this will remove all the '.' chars
There are times where I want calculations handled in double but I want the result displayed as as an int or even rounded amount, so the question isn't so strange (assuming that the given sample is simplified in order to ask the question).
I was going to post sample code for rounding, but it makes more sense to just use the built-in method Math.Round(). You can cast to a long, as mentioned above, but you won't have rounding, if desired (which it usually is, IMHO).
txtEditFormId.Text = ((long)formId).ToString();

Convert text data into percentage in C#

I am getting data into a text field and I need to display it as a percentage. Is there a function to perform this?
Ex: in my column I have "0.5", "0.1","0.2","0.25" etc., which needs to be displayed as
50%,10%,20%,25% etc., What is the best way to do it?
You should do this in two phases:
Parse the text as a number so you've got the value as your "real" type. (As a general rule, parse from text as early as you can, and format to a string as late as you can... operations between the two will be a lot simpler using the natural type.)
Format the number as a percentage using the standard numeric format string for percentage
So:
decimal percentage = decimal.Parse(input);
string output = percentage.ToString("p0");
Notes:
You should consider both input and output culture; are you always expecting to use "." as the decimal separator, for example?
Use decimal rather than double to exactly represent the value in the text (for example, the text could have "0.1" but double can't hold a value of exactly 0.1)
You can add things like desired precision to the formatting; see the linked docs for details; the example gives just an integer percentage, for example
Easiest would be to parse it (must be a double) then convert it back to a string, formatting it as a percentage.
var percentageString = double.Parse(doubleString).ToString("p1");
Now, some of you hoity-toity types may say that decimal is the correct type to use in this case.
Well, yes, if you need an additional 12-13 digits of precision.
However, most of us real folk (and I'm all about keeping it real) are fine with double's 15-16 digits of precision.
The real choice is whether or not your code is using doubles or decimals in the first place. If you are using doubles in your code, just stick with doubles. If decimals, stick to decimals. What you definitely do want to avoid is having to convert between the two any more than is absolutely necessary, as there be dragons. And unexpected runtime bugs that can corrupt your data. But mostly dragons.

C# Convert.ToDouble() loses decimal points when converting string to double

Let's say we have the following simple code
string number = "93389.429999999993";
double numberAsDouble = Convert.ToDouble(number);
Console.WriteLine(numberAsDouble);
after that conversion numberAsDouble variable has the value 93389.43. What can i do to make this variable keep the full number as is without rounding it? I have found that Convert.ToDecimal does not behave the same way but i need to have the value as double.
-------------------small update---------------------
putting a breakpoint in line 2 of the above code shows that the numberAsDouble variable has the rounded value 93389.43 before displayed in the console.
93389.429999999993 cannot be represented exactly as a 64-bit floating point number. A double can only hold 15 or 16 digits, while you have 17 digits. If you need that level of precision use a decimal instead.
(I know you say you need it as a double, but if you could explain why, there may be alternate solutions)
This is expected behavior.
A double can't represent every number exactly. This has nothing to do with the string conversion.
You can check it yourself:
Console.WriteLine(93389.429999999993);
This will print 93389.43.
The following also shows this:
Console.WriteLine(93389.429999999993 == 93389.43);
This prints True.
Keep in mind that there are two conversions going on here. First you're converting the string to a double, and then you're converting that double back into a string to display it.
You also need to consider that a double doesn't have infinite precision; depending on the string, some data may be lost due to the fact that a double doesn't have the capacity to store it.
When converting to a double it's not going to "round" any more than it has to. It will create the double that is closest to the number provided, given the capabilities of a double. When converting that double to a string it's much more likely that some information isn't kept.
See the following (in particular the first part of Michael Borgwardt's answer):
decimal vs double! - Which one should I use and when?
A double will not always keep the precision depending on the number you are trying to convert
If you need to be precise you will need to use decimal
This is a limit on the precision that a double can store. You can see this yourself by trying to convert 3389.429999999993 instead.
The double type has a finite precision of 64 bits, so a rounding error occurs when the real number is stored in the numberAsDouble variable.
A solution that would work for your example is to use the decimal type instead, which has 128 bit precision. However, the same problem arises with a smaller difference.
For arbitrary large numbers, the System.Numerics.BigInteger object from the .NET Framework 4.0 supports arbitrary precision for integers. However you will need a 3rd party library to use arbitrary large real numbers.
You could truncate the decimal places to the amount of digits you need, not exceeding double precision.
For instance, this will truncate to 5 decimal places, getting 93389.42999. Just replace 100000 for the needed value
string number = "93389.429999999993";
decimal numberAsDecimal = Convert.ToDecimal(number);
var numberAsDouble = ((double)((long)(numberAsDecimal * 100000.0m))) / 100000.0;

System.Single accuracy with C#

In the example below the number 12345678.9 loses accuracy when it's converted to a string as it becomes 1.234568E+07. I just need a way to preserve the accuracy for large floating point numbers. Thanks.
Single sin1 = 12345678.9F;
String str1 = sin1.ToString();
Console.WriteLine(str1); // displays 1.234568E+07
If you want to preserve decimal numbers, you should use System.Decimal. It's as simple as that. System.Single is worse than System.Double in that as per the documentation:
By default, a Single value contains only 7 decimal digits of precision, although a maximum of 9 digits is maintained internally.
You haven't just lost information when you've converted it to a string - you've lost information in the very first line. That's not just because you're using float instead of double - it's because you're using a floating binary point number.
The decimal number 0.1 can't be represented accurately in a binary floating point system no matter how big you make the type...
See my articles on floating binary point and floating decimal point for more information. Of course, it's possible that you should be using double or even float and just not caring about the loss of precision - it depends on what you're trying to represent. But if you really do care about preserving decimal digits, then use a decimal-based type.
You can't. Simple as that. In memory your number is 12345679. Try the code below.
Single sin1 = 12345678.9F;
String str1 = sin1.ToString("r"); // Shows "all" the number
Console.WriteLine(sin1 == 12345679); // true
Console.WriteLine(str1); // displays 12345679
Technically r means (quoting from MSDN) round-trip: Result: A string that can round-trip to an identical number. so in reality it isn't showing all the decimals. It's only showing all the decimals needed to distinguish it from other possible values of Single. If you want to show all the decimals use F20.
If you want more precision use double or better use decimal. float has the precision that it has. As we say in Italy "Non puoi spremere sangue da una rapa" (You can't squeeze blood from a turnip)
You could also write an IFormatProvider for your purpose - but the precision doesn't get any better unless you use a different type.
this article may help - http://www.csharp-examples.net/string-format-double/

Categories

Resources