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Double's signum should return 0 for -0d. #5766

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scabug opened this issue May 7, 2012 · 3 comments
Closed

Double's signum should return 0 for -0d. #5766

scabug opened this issue May 7, 2012 · 3 comments
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@dgruntz
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2.10.0-M2

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@scabug
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scabug commented May 7, 2012

Currently, (-0d).signum returns -1. I would expect this to return 0. For reference, the following works fine:

assert(scala.math.signum(-0d) == 0)
assert(java.lang.Math.signum(-0d) == 0)

But this blows up:

assert((-0d).signum == 0)

I am aware you want to follow Java's java.lang.Double.compare for Ordering (which strangely believes 0 * -1.0 < 0.0), but in this case I believe we should follow math/Math.signum's lead.

I think this would require overriding signum in DoubleIsConflicted from Numeric.scala.
@scabug
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scabug commented May 7, 2012

Imported From: https://issues.scala-lang.org/browse/SI-5766?orig=1
Reporter: Thomas Switzer (tixxit)
Affected Versions: 2.9.2
Other Milestones: 2.10.0-M3

@scabug
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scabug commented May 7, 2012

@dgruntz said:
This is fixed in the 2.10.0 M3:

scala> scala.math.signum(-0d)
res0: Double = 0.0

scala> java.lang.Math.signum(-0d)
res1: Double = -0.0

scala> (-0d).signum
res2: Int = 0

@scabug scabug closed this as completed May 8, 2012
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scabug commented May 8, 2012

@dgruntz said:
Actually, in the current head release scala.math.signum(-0d) returns a negative zero (note, that a negative zero is equal to a positive zero if compared with ==).

scala> math.signum(-0d)
res0: Double = -0.0

scala> res0 == 0
res1: Boolean = true

@scabug scabug added this to the 2.10.0-M2 milestone Apr 7, 2017
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