How to get remainder of 2 double using % operator?

Question

I am trying to get remainder of 2 double using % operator

double x = 965.606698735538;
double y = 482.803349367769;
var mod = x % y;

Expected value is 0

Actual value is 482.803349367769

Can someone help why its behaving so strange.

@scartag in the OP's case, `2x == y` so the answer should be zero. — Dai, Sep 12 '17 at 09:43
`Console.WriteLine(x % y)` returns 0 in my fiddle. What's going wrong actually? — Tetsuya Yamamoto, Sep 12 '17 at 09:43
Most likely you're a victim of floating point limited accuracy/precision: http://floating-point-gui.de/ — Crozin, Sep 12 '17 at 09:43
Like @TetsuyaYamamoto says: This actually works, so the error must be elsewhere. It'll be a rounding error. — Matthew Watson, Sep 12 '17 at 09:44
I intentionally added one decimal place on the second operand & suddenly it turns into `482.xxxx`. The `double` data type precision error exactly confirmed the issue. — Tetsuya Yamamoto, Sep 12 '17 at 09:46
Though the duplicate question deals with + the answer answers perfectly why this happens, floating point is not accurate so the division isn't perfectly divisable by 2. — Lasse V. Karlsen, Sep 12 '17 at 09:47

Bathsheba · Answer 1 · 2017-09-12T09:46:09.197

3

This is a quirk of binary floating point arithmetic.

The closest (IEEE754) double to 482.803349367769 is 482.80334936776898757671006023883819580078125

The closest double to 965.606698735538 is 965.6066987355379751534201204776763916015625

As you can see the latter is not exactly twice the former; in fact it is just under twice that. Which accounts for the large remainder that you see.

You need to be careful when using % with floating point arguments.

edited Sep 12 '17 at 09:46

answered Sep 12 '17 at 09:44

Bathsheba

This is true, but if you actually run the OP's code, it returns 0. – Matthew Watson Sep 12 '17 at 09:45
@MatthewWatson: That is the *formatter* being clever with the output. – Bathsheba Sep 12 '17 at 09:46
1

It's not the formatter - there's no way a formatter would change ~482 to 0... – Matthew Watson Sep 12 '17 at 09:47
@MatthewWatson: Touche. – Bathsheba Sep 12 '17 at 09:48
But it definitely is being caused by a rounding error, like you say - it's just that the OP's code does not actually exhibit the problem (at least, not on my PC). – Matthew Watson Sep 12 '17 at 09:49
I assume it is the formatting of the numbers that the OP is reading out of the debugger. The ones in the question are almost certainly calculated in his original code. – Chris Sep 12 '17 at 09:49
@MatthewWatson: It's possible their processor is using 80 bit floating point internally. – Bathsheba Sep 12 '17 at 09:49
@Bathsheba Do you have any alternate solution – paresh Sep 12 '17 at 12:39
C# has a built-in decimal type which could be a good alternative. Personally I'd arrange the calculations so the remainder operator takes integral type arguments. – Bathsheba Sep 12 '17 at 12:43
I use the aproach of calculating the rest missing to the next or previous "full number" and having a tolerance value. See this: https://gist.github.com/GW-FUB/1a5d3f3abb88aba90aec6de0ea461d53 – Gabriel Weidmann Nov 06 '20 at 07:15

1 Answers1