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%28double
29%。
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::expm1
和std::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。
例
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
exp | returns 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
© cppreference.com
在CreativeCommonsAttribution下授权-ShareAlike未移植许可v3.0。