int feclearexcept( int excepts );
(since C99)Attempts to clear the floating-point exceptions that are listed in the bitmask argument excepts, which is a bitwise OR of the floating-point exception macros.
â0â if all indicated exceptions were successfully cleared or if excepts is zero. Returns a non-zero value on error.
#include <fenv.h>
#include <stdio.h>
#include <math.h>
#include <float.h>
/*
* A possible implementation of hypot which makes use of many advanced
* floating-point features.
*/
double hypot_demo(double a, double b) {
const int range_problem = FE_OVERFLOW | FE_UNDERFLOW;
feclearexcept(range_problem);
// try a fast algorithm
double result = sqrt(a * a + b * b);
if (!fetestexcept(range_problem)) // no overflow or underflow
return result; // return the fast result
// do a more complicated calculation to avoid overflow or underflow
int a_exponent,b_exponent;
frexp(a, &a_exponent);
frexp(b, &b_exponent);
if (a_exponent - b_exponent > DBL_MAX_EXP)
return fabs(a) + fabs(b); // we can ignore the smaller value
// scale so that fabs(a) is near 1
double a_scaled = scalbn(a, -a_exponent);
double b_scaled = scalbn(b, -a_exponent);
// overflow and underflow is now impossible
result = sqrt(a_scaled * a_scaled + b_scaled * b_scaled);
// undo scaling
return scalbn(result, a_exponent);
}
int main(void)
{
// Normal case takes the fast route
printf("hypot(%f, %f) = %f\n", 3.0, 4.0, hypot_demo(3.0, 4.0));
// Extreme case takes the slow but more accurate route
printf("hypot(%e, %e) = %e\n", DBL_MAX / 2.0,
DBL_MAX / 2.0,
hypot_demo(DBL_MAX / 2.0, DBL_MAX / 2.0));
return 0;
}
Output:
hypot(3.000000, 4.000000) = 5.000000 hypot(8.988466e+307, 8.988466e+307) = 1.271161e+308[edit] References
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