8 replies to this topic

[JambaFun]

Hey,

Please explain this:

Mod(10,1) -> 0
Mod(1.0,0.1) -> 0.1

It seems like a bug to me...


Cheers!

[SKAN]

Use Round()

If Round( Mod(1.0,0.1) ) 
   MsgBox

[JambaFun]

Tsss...
No can do since I want the floating point remainder.

[Lexikos]

What floating point remainder?

Mod(Dividend, Divisor): Modulo. Returns the remainder when Dividend is divided by Divisor.

Mod(1.0,0.1)

1.0 / 0.1 = 10

No remainder...

Btw, you can redefine built-in functions.

mod(a, b) {
  return a - b * floor(a/b)
}

Obviously not as efficient as what AHK uses (qmathFmod(); 6 assembly instructions :shock:.)

[JambaFun]

What floating point remainder?

Well, zero of course.

Mod(1.0,0.1) incorrectly returns 0.1.

[Laszlo]

This could be the old inexact floating point representation problem. Short decimal numbers have often infinite, periodic binary representations, which have to be truncated internally. Consequently, the results of arithmetic operations are also just approximate.

SetFormat Float, 0.20

MsgBox % Mod(1.0,0.1) ; 0.09999999999999995000

There is no bug here. You cannot always expect exact floating point results.

[Raccoon]

Strange. I don't know what to say about this.

Clearly, Mod(10,1) = 0, so Mod(1.0,0.1) should also result in 0. And yet, 3 other languages I've tested this in returns Mod(1.0,0.1) = 0.1, while most calculators return 0.

Consider Mod(0.3,0.2) should be 0.1, no? And that's what I get in those languages that support non-integer values, but Mod(1.0,0.1) just doesn't make sense for it to return 0.1. It's a mystery to me.

I'm going to ask in some #math IRC channels.

[Laszlo]

Read the posts above: you are asking the exact ratio of inexact numbers. If you perform fixpoint arithmetic, or BCD, your result will be different. Floats often are inexact, in any language.