> On Nov 9, 2017, at 10:55 AM, Rick Mann via swift-users
> <swift-users@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.
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