Double ToString format with no precision lost and group separators

Question

I am trying to get the string output of a Double value with group separators and no loss in precision in any culture. Is there anyway to do this without writing my own method to do it?

I know you can use ToString("N0"), but this results in a loss of precision. I know you can use ToString("R") to keep the precision, but this doesn't include group separators.

E.g. For Double.MaxValue in Invariant Culture I would like

"179,769,313,486,231,570,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000"

Edit: When I say loss in precision I mean the full number of significant figures rather than not loosing precision in the 'decmial' part of the value. I want to be able to do 'round trips' with group separators, so ToString -> Parse -> ToString.

Edit 2: For a better comparison of the problem I'm trying to solve here is the output when using ToString("N0"). Here you can see the rounding at the end of the significant figures:

"179,769,313,486,232,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000"

http://stackoverflow.com/questions/2863388/what-is-the-equivalent-of-the-java-bigdecimal-class-in-c — Aurimas Neverauskas, Nov 20 '14 at 16:51
[`decimal`](http://msdn.microsoft.com/en-us/library/364x0z75.aspx) — Scott Solmer, Nov 20 '14 at 16:52
I've edited to explain my use of the word precision above. This is not about decimal precision, but about loss of significant figures in the string. — MarkOwen320, Nov 20 '14 at 17:06
The question is what do you need this for? It seems like an odd requirement. — Magnus, Nov 20 '14 at 22:36
In my case it is to generate values to test a Double parser in multiple different cultures, but I could definitely see someone wanting this for other reasons. For example I can see someone wanting to output the Single/float "12345678" with group separators without losing the significant figures, but that comes out as 12,345,680 with ToString("N0"). This is the same issue as described above. — MarkOwen320, Nov 20 '14 at 22:53
@Magnus "r" or "R" (I believe they are the same) doesn't give the group separators — MarkOwen320, Nov 21 '14 at 09:42
I think it is the best you're gonna get. You can add the group separation yourself afterwards. — Magnus, Nov 21 '14 at 09:56
That's what I thought when I posted, but wanted to check if anyone had any alternatives. Thanks — MarkOwen320, Nov 21 '14 at 15:17

score 1 · Accepted Answer · answered Dec 01 '14 at 11:56

Doesn't look like any of the built in ToString formats do what I required.

For my purposes I was able to write a method to deal with the scenarios I had, but since it doesn't work for all possible values I'm not posting it here.

Thanks to all the posters for their assistance.

score 0 · Answer 2 · answered Nov 20 '14 at 17:03

0

If you specify the format #,0.# you will get:

Thousand separators
As many decimals as required

That should fulfill your requirement, if I'm reading the question correctly.

answered Nov 20 '14 at 17:03

Bas

26,772
8
53
86

I'm afraid that doesn't work as you loose the significant figures. In the example below you will notice rounding at the '232,000' part, rather than the '231,570' in my original question: Double.MaxValue.ToString("#,0.#") = "179,769,313,486,232,000,000..." – MarkOwen320 Nov 20 '14 at 20:28

Double ToString format with no precision lost and group separators

2 Answers2