1 files changed, 0 insertions, 97 deletions
diff --git a/src/port/rint.c b/src/port/rint.c
deleted file mode 100644
index d59d9ab7743..00000000000
--- a/src/port/rint.c
+++ /dev/null
@@ -1,97 +0,0 @@
-/*-------------------------------------------------------------------------
- *
- * rint.c
- *	  rint() implementation
- *
- * By Pedro Gimeno Fortea, donated to the public domain
- *
- * IDENTIFICATION
- *	  src/port/rint.c
- *
- *-------------------------------------------------------------------------
- */
-#include "c.h"
-
-#include <math.h>
-
-/*
- * Round to nearest integer, with halfway cases going to the nearest even.
- */
-double
-rint(double x)
-{
-	double		x_orig;
-	double		r;
-
-	/* Per POSIX, NaNs must be returned unchanged. */
-	if (isnan(x))
-		return x;
-
-	if (x <= 0.0)
-	{
-		/* Both positive and negative zero should be returned unchanged. */
-		if (x == 0.0)
-			return x;
-
-		/*
-		 * Subtracting 0.5 from a number very close to -0.5 can round to
-		 * exactly -1.0, producing incorrect results, so we take the opposite
-		 * approach: add 0.5 to the negative number, so that it goes closer to
-		 * zero (or at most to +0.5, which is dealt with next), avoiding the
-		 * precision issue.
-		 */
-		x_orig = x;
-		x += 0.5;
-
-		/*
-		 * Be careful to return minus zero when input+0.5 >= 0, as that's what
-		 * rint() should return with negative input.
-		 */
-		if (x >= 0.0)
-			return -0.0;
-
-		/*
-		 * For very big numbers the input may have no decimals.  That case is
-		 * detected by testing x+0.5 == x+1.0; if that happens, the input is
-		 * returned unchanged.  This also covers the case of minus infinity.
-		 */
-		if (x == x_orig + 1.0)
-			return x_orig;
-
-		/* Otherwise produce a rounded estimate. */
-		r = floor(x);
-
-		/*
-		 * If the rounding did not produce exactly input+0.5 then we're done.
-		 */
-		if (r != x)
-			return r;
-
-		/*
-		 * The original fractional part was exactly 0.5 (since
-		 * floor(input+0.5) == input+0.5).  We need to round to nearest even.
-		 * Dividing input+0.5 by 2, taking the floor and multiplying by 2
-		 * yields the closest even number.  This part assumes that division by
-		 * 2 is exact, which should be OK because underflow is impossible
-		 * here: x is an integer.
-		 */
-		return floor(x * 0.5) * 2.0;
-	}
-	else
-	{
-		/*
-		 * The positive case is similar but with signs inverted and using
-		 * ceil() instead of floor().
-		 */
-		x_orig = x;
-		x -= 0.5;
-		if (x <= 0.0)
-			return 0.0;
-		if (x == x_orig - 1.0)
-			return x_orig;
-		r = ceil(x);
-		if (r != x)
-			return r;
-		return ceil(x * 0.5) * 2.0;
-	}
-}