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Equivalent of the php fmod in C#

Answer 1

A:

Have you tried the % modulus operator?

AaronLS 2009-11-20 18:35:05

fmod is totally different.

plinth 2009-11-20 18:37:31

@plinth: "fmod — Returns the floating point remainder (modulo) of the division of the arguments" -- PHP Manual

R. Bemrose 2009-11-20 18:41:09

@plinth "The modulus operator (%) computes the remainder after dividing its first operand by its second." -- C# manual

R. Bemrose 2009-11-20 18:42:35

fmod works on floating point numbers - the fmod(10.0, 3.1) is different from (double)((int)10.0 % (int)3.1). "The fmod function calculates the floating-point remainder f of x / y such that x = i * y + f, where i is an integer, f has the same sign as x, and the absolute value of f is less than the absolute value of y."

plinth 2009-11-20 18:45:06

@R. Bemrose, they're not the same. `fmod(5.7, 1.3)` gives `0.5`, while `5.7 % 1.3` gives `0`.

Ionuț G. Stan 2009-11-20 18:45:48

@Ionut G. Stan: `double test = 5.7 % 1.3; Console.WriteLine(test.ToString();` prints `0.5`

R. Bemrose 2009-11-20 18:50:15

@R. Bemrose. You are right. In c# it behaves differently.

Ionuț G. Stan 2009-11-20 18:53:24

or rather, it would had I not omitted one of the paratheses (whoops)

R. Bemrose 2009-11-20 18:54:31

Stan, you may have a casting issue. Because of the way C# will coerce types during arithmetic, sometimes you have to explicitely cast them and/or specify the types of numeric literals. Can you give a more complete code for how you tested "5.7 % 1.3 gives 0"

AaronLS 2009-11-20 18:54:41

I stand corrected - msdn says that modulus is defined for all numeric types, and % for doubles *is* fmod.

plinth 2009-11-20 19:05:00

Answer 2

+2 A:

Use a P/Invoked call into the standard library (from here):

[DllImport("msvcrt.dll")]
static extern double fmod(double x, double y);

plinth 2009-11-20 18:37:03

That works but I was hoping for a C# solution without having to import a DLL.

John 2009-11-20 18:41:58

You will always have that dll.

plinth 2009-11-20 18:46:03

CORRECTION: do not use this. Use the built-in modulus operator, it works correctly for all numeric types. Pinvoking to fmodf will add unnecessary overhead.

plinth 2009-11-20 19:06:02

I changed the accepted answer to the one that uses the modulus operator. Thanks all.

John 2009-11-20 21:00:04

"You will always have `msvcrt.dll`". Anyone heard of Mono? Silverlight?

MarkJ 2009-11-20 21:07:04

Answer 3

+3 A:

Other people have given the correct operator to use, but not the entire syntax, which would be:

double lon_rad = ((lon1+dlon_rad + Math.PI) % (2*Math.PI)) - Math.PI;

R. Bemrose 2009-11-20 18:45:04

I would change the "2" to "2.0"

AaronLS 2009-11-20 18:56:12

That's probably a good idea for optimization, but I'm sure that it would affect the results, as Math.PI is already a double.

R. Bemrose 2009-11-20 19:17:35

True. I knew PI was a double, but I can never remember which direction it will coerce the operands whenever they are of different types so I eliminate any doubt by clarifying the numeric literals.

AaronLS 2009-11-20 19:35:31

It always makes them the larger of the two types. It makes sense if you think about similar types. For instance, if I multiply a byte by an int, the result will be an int because there are int values outside the byte range.

R. Bemrose 2009-11-20 19:56:55

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Equivalent of the php fmod in C#

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