1 /*-------------------------------------------------------------------------
4 * rint() implementation
6 * By Pedro Gimeno Fortea, donated to the public domain
11 *-------------------------------------------------------------------------
19 * Round to nearest integer, with halfway cases going to the nearest even.
27 /* Per POSIX, NaNs must be returned unchanged. */
33 /* Both positive and negative zero should be returned unchanged. */
38 * Subtracting 0.5 from a number very close to -0.5 can round to
39 * exactly -1.0, producing incorrect results, so we take the opposite
40 * approach: add 0.5 to the negative number, so that it goes closer to
41 * zero (or at most to +0.5, which is dealt with next), avoiding the
48 * Be careful to return minus zero when input+0.5 >= 0, as that's what
49 * rint() should return with negative input.
55 * For very big numbers the input may have no decimals. That case is
56 * detected by testing x+0.5 == x+1.0; if that happens, the input is
57 * returned unchanged. This also covers the case of minus infinity.
59 if (x == x_orig + 1.0)
62 /* Otherwise produce a rounded estimate. */
66 * If the rounding did not produce exactly input+0.5 then we're done.
72 * The original fractional part was exactly 0.5 (since
73 * floor(input+0.5) == input+0.5). We need to round to nearest even.
74 * Dividing input+0.5 by 2, taking the floor and multiplying by 2
75 * yields the closest even number. This part assumes that division by
76 * 2 is exact, which should be OK because underflow is impossible
77 * here: x is an integer.
79 return floor(x * 0.5) * 2.0;
84 * The positive case is similar but with signs inverted and using
85 * ceil() instead of floor().
91 if (x == x_orig - 1.0)
96 return ceil(x * 0.5) * 2.0;