C
数值 | Numerics

feclearexcept

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。

#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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