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数字 | Numerics

std::expm1

性病:费用1

Defined in header
float expm1( float arg (1)(since C++11)
double expm1( double arg (2)(since C++11)
long double expm1( long double arg (3)(since C++11)
double expm1( Integral arg (4)(since C++11)

1-3%29计算e%28欧拉%27S数,2.7182818%29提高到给定的功率arg,减1.0这个函数比表达式更精确std::exp(arg)-1.0如果arg接近于零。

4%29一组过载或接受任意参数的函数模板积分型等于2%29%28double29%。

参数

arg-value of floating-point or Integral type

返回值

如果没有错误发生

-1人被退回。

如果溢出导致范围错误,+HUGE_VAL,,,+HUGE_VALF,或+HUGE_VALL会被归还。

如果由于下流而发生范围错误,则返回舍入%29后的正确结果%28。

错误处理

错误按数学[医]错误处理...

如果实现支持ieee浮点算法%28IEC 60559%29,

  • 如果参数为±0,则返回未经修改的参数

  • 如果参数为-∞,则返回-1。

  • 如果参数为+∞,则返回+∞

  • 如果参数为nan,则返回nan。

注记

功能std::expm1std::log1p用于财务计算,例如,在计算小日利率时:%281+x%29N

-1可以表示为std::expm1(n *std::log1p(x))这些函数还简化了精确逆双曲函数的编写。

适用于ieee兼容的类型。double,如果709.8<arg,则保证溢出。

二次

#include <iostream> #include <cmath> #include <cerrno> #include <cstring> #include <cfenv> #pragma STDC FENV_ACCESS ON int main() { std::cout << "expm1(1) = " << std::expm1(1) << '\n' << "Interest earned in 2 days on on $100, compounded daily at 1%\n" << " on a 30/360 calendar = " << 100*std::expm1(2*std::log1p(0.01/360)) << '\n' << "exp(1e-16)-1 = " << std::exp(1e-16)-1 << ", but expm1(1e-16) = " << std::expm1(1e-16) << '\n'; // special values std::cout << "expm1(-0) = " << std::expm1(-0.0) << '\n' << "expm1(-Inf) = " << std::expm1(-INFINITY) << '\n'; // error handling errno=0; std::feclearexcept(FE_ALL_EXCEPT std::cout << "expm1(710) = " << std::expm1(710) << '\n'; if(errno == ERANGE) std::cout << " errno == ERANGE: " << std::strerror(errno) << '\n'; if(std::fetestexcept(FE_OVERFLOW)) std::cout << " FE_OVERFLOW raised\n"; }

二次

可能的产出:

二次

expm1(1) = 1.71828 Interest earned in 2 days on on $100, compounded daily at 1% on a 30/360 calendar = 0.00555563 exp(1e-16)-1 = 0 expm1(1e-16) = 1e-16 expm1(-0) = -0 expm1(-Inf) = -1 expm1(710) = inf errno == ERANGE: Result too large FE_OVERFLOW raised

二次

另见

expreturns e raised to the given power (ex) (function)
exp2 (C++11)returns 2 raised to the given power (2x) (function)
log1p (C++11)natural logarithm (to base e) of 1 plus the given number (ln(1+x)) (function)

c费用文件1

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