Here's a possibly faster implementation similar to @Roddy's answer:
typedef int64_t i_t;
typedef double f_t;
static inline
i_t i_tmin(i_t x, i_t y) {
return (y + ((x - y) & -(x < y))); // min(x, y)
}
static inline
i_t i_tmax(i_t x, i_t y) {
return (x - ((x - y) & -(x < y))); // max(x, y)
}
f_t clip_f_t(f_t f, f_t fmin, f_t fmax)
{
#ifndef TERNARY
assert(sizeof(i_t) == sizeof(f_t));
//assert(not (fmin < 0 and (f < 0 or is_negative_zero(f))));
//XXX assume IEEE-754 compliant system (lexicographically ordered floats)
//XXX break strict-aliasing rules
const i_t imin = *(i_t*)&fmin;
const i_t imax = *(i_t*)&fmax;
const i_t i = *(i_t*)&f;
const i_t iclipped = i_tmin(imax, i_tmax(i, imin));
#ifndef INT_TERNARY
return *(f_t *)&iclipped;
#else /* INT_TERNARY */
return i < imin ? fmin : (i > imax ? fmax : f);
#endif /* INT_TERNARY */
#else /* TERNARY */
return fmin > f ? fmin : (fmax < f ? fmax : f);
#endif /* TERNARY */
}
See Compute the minimum (min) or maximum (max) of two integers without branching and Comparing floating point numbers
The IEEE float and double formats were
designed so that the numbers are
“lexicographically ordered”, which –
in the words of IEEE architect William
Kahan means “if two floating-point
numbers in the same format are ordered
( say x < y ), then they are ordered
the same way when their bits are
reinterpreted as Sign-Magnitude
integers.”
A test program:
/** gcc -std=c99 -fno-strict-aliasing -O2 -lm -Wall *.c -o clip_double && clip_double */
#include <assert.h>
#include <iso646.h> // not, and
#include <math.h> // isnan()
#include <stdbool.h> // bool
#include <stdint.h> // int64_t
#include <stdio.h>
static
bool is_negative_zero(f_t x)
{
return x == 0 and 1/x < 0;
}
static inline
f_t range(f_t low, f_t f, f_t hi)
{
return fmax(low, fmin(f, hi));
}
static const f_t END = 0./0.;
#define TOSTR(f, fmin, fmax, ff) ((f) == (fmin) ? "min" : \
((f) == (fmax) ? "max" : \
(is_negative_zero(ff) ? "-0.": \
((f) == (ff) ? "f" : #f))))
static int test(f_t p[], f_t fmin, f_t fmax, f_t (*fun)(f_t, f_t, f_t))
{
assert(isnan(END));
int failed_count = 0;
for ( ; ; ++p) {
const f_t clipped = fun(*p, fmin, fmax), expected = range(fmin, *p, fmax);
if(clipped != expected and not (isnan(clipped) and isnan(expected))) {
failed_count++;
fprintf(stderr, "error: got: %s, expected: %s\t(min=%g, max=%g, f=%g)\n",
TOSTR(clipped, fmin, fmax, *p),
TOSTR(expected, fmin, fmax, *p), fmin, fmax, *p);
}
if (isnan(*p))
break;
}
return failed_count;
}
int main(void)
{
int failed_count = 0;
f_t arr[] = { -0., -1./0., 0., 1./0., 1., -1., 2,
2.1, -2.1, -0.1, END};
f_t minmax[][2] = { -1, 1, // min, max
0, 2, };
for (int i = 0; i < (sizeof(minmax) / sizeof(*minmax)); ++i)
failed_count += test(arr, minmax[i][0], minmax[i][1], clip_f_t);
return failed_count & 0xFF;
}
In console:
$ gcc -std=c99 -fno-strict-aliasing -O2 -lm *.c -o clip_double && ./clip_double
It prints:
error: got: min, expected: -0. (min=-1, max=1, f=0)
error: got: f, expected: min (min=-1, max=1, f=-1.#INF)
error: got: f, expected: min (min=-1, max=1, f=-2.1)
error: got: min, expected: f (min=-1, max=1, f=-0.1)
a
to be returned exactly asa
, then many answers do not meet that hurdle. If precision is of no concern , then always returning(MY_MAX + MY_MIN)/2
will certainly be a fast low-precision answer, and certainly foolish. Recommend tolerating no more than 1 ULP error. – Stipulation__m128
)? – Pt