insaf021 commented on code in PR #349:
URL: https://github.com/apache/commons-math/pull/349#discussion_r4487264084
##########
commons-math-core/src/main/java/org/apache/commons/math4/core/jdkmath/AccurateMath.java:
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@@ -3710,9 +3710,20 @@ public static double IEEEremainder(final double
dividend, final double divisor)
}
} else {
// we are in the general case
- final double n = AccurateMath.rint(dividend / divisor);
- final double remainder = Double.isInfinite(n) ? 0.0 : dividend -
divisor * n;
- return (remainder == 0) ? AccurateMath.copySign(remainder,
dividend) : remainder;
+ // Reduce |dividend| modulo 2 |divisor| with the exact truncating
remainder
+ // (x is unchanged if 2 |divisor| overflows), then subtract
|divisor| at most
+ // twice to round the quotient to even, as in fdlibm
__ieee754_remainder.
+ // All steps are exact: x + x only overflows when x + x > p holds
anyway.
+ final double p = Math.abs(divisor);
+ double x = Math.abs(dividend) % (p + p);
Review Comment:
Yes, it's the same value. Doubling a double is exact, and 2 * p overflows to
infinity for exactly the same p as p + p, so the overflow path where % leaves x
unchanged behaves as before. Switched in 319201c and AccurateMathTest still
passes. The two x + x comparisons below are the same case if you'd prefer those
as 2 * x as well.
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