feclearexcept

在头文件中定义
int feclearexcept（int excepts）;		（自C99以来）

尝试清除位掩码参数excepts中列出的浮点异常，这是浮点异常宏的按位或的结果。

参数

excepts	-	bitmask listing the exception flags to clear

返回值

如果所有指示的例外都被成功清除或者除外为零，则为0。返回错误时的非零值。

例

#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;
}

输出：

hypot(3.000000, 4.000000) = 5.000000
hypot(8.988466e+307, 8.988466e+307) = 1.271161e+308

参考

C11标准（ISO / IEC 9899：2011）：

扩展内容

fetestexcept（C99）	确定哪个指定的浮点状态标志被设置（功能）

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