fi
-ac_fn_c_check_func "$LINENO" "rint" "ac_cv_func_rint"
-if test "x$ac_cv_func_rint" = xyes; then :
- $as_echo "#define HAVE_RINT 1" >>confdefs.h
-
-else
- case " $LIBOBJS " in
- *" rint.$ac_objext "* ) ;;
- *) LIBOBJS="$LIBOBJS rint.$ac_objext"
- ;;
-esac
-
-fi
-
ac_fn_c_check_func "$LINENO" "srandom" "ac_cv_func_srandom"
if test "x$ac_cv_func_srandom" = xyes; then :
$as_echo "#define HAVE_SRANDOM 1" >>confdefs.h
+++ /dev/null
-/*-------------------------------------------------------------------------
- *
- * 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;
- }
-}
HAVE_READLINE_HISTORY_H => undef,
HAVE_READLINE_READLINE_H => undef,
HAVE_READLINK => undef,
- HAVE_RINT => 1,
HAVE_RL_COMPLETION_APPEND_CHARACTER => undef,
HAVE_RL_COMPLETION_MATCHES => undef,
HAVE_RL_COMPLETION_SUPPRESS_QUOTE => undef,