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在头文件<fenv.h>中定义 | ||
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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。
excepts | - | bitmask listing the exception flags to clear |
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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) = 5.000000hypot(8.988466e+307, 8.988466e+307) = 1.271161e+308
C11标准(ISO / IEC 9899:2011):
7.6.2.1 feclearexcept函数(p:209)
C99标准(ISO / IEC 9899:1999):
7.6.2.1 feclearexcept函数(p:190)
fetestexcept(C99) | 确定哪个指定的浮点状态标志被设置(功能) |
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| 用于feclearexcept的C ++文档 |