Include trailing zeroes after rounding off a decimal/double value

Question

I am using Math.Round() Method to round off a double value. But I want to keep the trailing zeroes. For Example:

double x = 0.4701589;
Math.Round(x*100,2);

This returns 4.7 however the desired output is 4.70

Is `value` `double` or `decimal`? `double` doesn't retain any precision information. — Damien_The_Unbeliever, Feb 22 '21 at 10:43
You are getting `4.7` it's just equivalent to`4.70000000`. It's not about rounding but display. — Drag and Drop, Feb 22 '21 at 10:44
Does this answer your question? [How to display values only upto 2 decimal places](https://stackoverflow.com/questions/15288134/how-to-display-values-only-upto-2-decimal-places) — Drag and Drop, Feb 22 '21 at 10:45
Then the answer is no. The `double` type is incapable of storing a distinction between `4.7` and `4.70`. — Damien_The_Unbeliever, Feb 22 '21 at 10:46
In math, there's no difference between 4.7 and 4.70. This is a display issue. — Panagiotis Kanavos, Feb 22 '21 at 10:50
Then, is there a way, we can convert this double to decimal with 2 decimal points? — murli, Feb 22 '21 at 10:52
Not sure what the problem is, for numeric types there is no difference between 4.7 or 4.70 or 4.7000000000000. Are you wanting to display this value to 2 decimal places as a string? — phuzi, Feb 22 '21 at 11:00
i want to pass this as decimal in api response with 2 decimal places — murli, Feb 22 '21 at 11:03
Math.Round(Convert.ToDecimal(value*100), 2) will give correct results? — murli, Feb 22 '21 at 11:04
@murli: if you pass it in an api response, there is no difference between 4.7 and 4.70. On the receiving end both will be deserialized to the same value. — Johan Donne, Feb 22 '21 at 11:14
Have you considered using `decimal` rather than `double`? https://dotnetfiddle.net/ReWIuo — mjwills, Feb 22 '21 at 11:41
@murli `Math.Round` won't do that without specifying a midpoint rounding parameter. By default it will be 4.6 / 4.60 (bankers' rounding). A common mistake with `Math.Round`. — mjwills, Feb 22 '21 at 11:42

score 2 · Answer 1 · answered Feb 22 '21 at 10:48

The problem is not with the rounding but with the formatting of your result. You can specify it needs to be printed with 2 decimals:

        double x = 0.04701589;
        double y = Math.Round(x*100, 2);

        Console.WriteLine($"default formatting: {y}");
        Console.WriteLine($"specify 2 digits: {y:f2}");

AmirJabari · Answer 2 · 2021-02-22T11:53:22.370

0

You can use .ToString("0.00") whenever you want to display it

Console.WriteLine(finalValue.ToString("0.00"));

edited Feb 22 '21 at 11:53

answered Feb 22 '21 at 10:52

AmirJabari

406
3
9

This prints `4.7`, not `4.70`. Also after your update. – Peter B Feb 22 '21 at 10:55
This does change the internal representation of the `decimal`. It's a valid answer. – Enigmativity Feb 22 '21 at 11:00
@Enigmativity What makes it valid? `Console.WriteLine(x);` produces `4.7` – Peter B Feb 22 '21 at 11:08
@AmirJabari Have you even tried it? I guess not, because what you say **does not produce `4.70`**. – Peter B Feb 22 '21 at 11:27
1

It doesn't work at https://dotnetfiddle.net/TfqJBb . – mjwills Feb 22 '21 at 11:48
Yes, I now admit my mistake. '0' not '#'. I learned something new. – AmirJabari Feb 22 '21 at 12:07

score 0 · Answer 3 · answered Feb 22 '21 at 11:32

Try this

int numDigitsAfterPoint = 5;
double num = 1.25d;
string result = num.ToString("F" + numDigitsAfterPoint);

Keep in mind that, ToString uses the MidpointRounding.AwayFromZero instead of the MidpointRounding.ToEven (also called Banker's Rounding). As an example:

var x1 = 1.25.ToString("F1");
var x2 = 1.35.ToString("F1");
var x3 = Math.Round(1.25, 1).ToString();
var x4 = Math.Round(1.35, 1).ToString();

These will produce different result (because Math.Round normally uses MidpointRounding.ToEven)

Include trailing zeroes after rounding off a decimal/double value

3 Answers3