[swift-users] Modulo operation in Swift?

[Stephen Canon]

Use fmod(x,y). It is identical to C fmod(x,y). It’s also identical to x.truncatingRemainder(dividingBy: y).

> On Nov 9, 2017, at 1:12 PM, Roderick Mann <rmann at latencyzero.com> wrote:
> 
> Ah, I see.
> 
> So, here's my issue. I recently ported some C code for satellite position prediction to Swift. This code used fmod() a lot. I had thought fmod() was just % for floats, but I see now it's not.
> 
> What is truncatingRemainder in all this? It would be great if there were a documentation section covering this in detail (in the context of Swift). It's confusing because fmod() is available, but so are the other options.
> 
> Should %% be something the language provides for IEEE 754 remainder()?
> 
>> On Nov 9, 2017, at 11:18 , Stephen Canon <scanon at apple.com> wrote:
>> 
>>> On Nov 9, 2017, at 10:55 AM, Rick Mann via swift-users <swift-users at swift.org> wrote:
>>> 
>>> Why is % not available for floating point numbers?
>> 
>> This has been discussed extensively. Swift had this this operator originally, but we removed it. Here’s the rationale I gave back then:
>> 
>> -----------
>> 
>> While C and C++ do not provide the “%” operator for floating-point types, many newer languages do (Java, C#, and Python, to name just a few).  Superficially this seems reasonable, but there are severe gotchas when % is applied to floating-point data, and the results are often extremely surprising to unwary users.  C and C++ omitted this operator for good reason.  Even if you think you want this operator, it is probably doing the wrong thing in subtle ways that will cause trouble for you in the future.
>> 
>> The % operator on integer types satisfies the division algorithm axiom: If b is non-zero and q = a/b, r = a%b, then a = q*b + r.  This property does not hold for floating-point types, because a/b does not produce an integral value.  If it did produce an integral value, it would need to be a bignum type of some sort (the integral part of DBL_MAX / DBL_MIN, for example, has over 2000 bits or 600 decimal digits).
>> 
>> Even if a bignum type were returned, or if we ignore the loss of the division algorithm axiom, % would still be deeply flawed.  Whereas people are generally used to modest rounding errors in floating-point arithmetic, because % is not continuous small errors are frequently enormously magnified with catastrophic results:
>> 
>> 	(swift) 10.0 % 0.1
>>    // r0 : Double = 0.0999999999999995 // What?!
>> 
>> [Explanation: 0.1 cannot be exactly represented in binary floating point; the actual value of “0.1” is 0.1000000000000000055511151231257827021181583404541015625.  Other than that rounding, the entire computation is exact.]
>> 
>> Proposed Approach:
>> Remove the “%” operator for floating-point types.
>> 
>> Alternative Considered:
>> Instead of binding “%” to fmod( ), it could be bound to remainder( ), which implements the IEEE 754 remainder operation; this is just like fmod( ), except instead of returning the remainder under truncating division, it returns the remainder of round-to-nearest division, meaning that if a and b are positive, remainder(a,b) is in the range [-b/2, b/2] rather than [0, b).  This still has a large discontinuity, but the discontinuity is moved away from zero, which makes it much less troublesome (that’s why IEEE 754 standardized this operation):
>> 
>> 	(swift) remainder(1, 0.1)
>>    // r1 : Double = -0.000000000000000055511151231257827 // Looks like normal floating-point rounding
>> 
>> The downside to this alternative is that now % behaves totally differently for integer and floating-point data, and of course the division algorithm still doesn’t hold.
>> 
>> 
> 
> 
> -- 
> Rick Mann
> rmann at latencyzero.com
> 
>