table of contents
std::expm1,std::expm1f,std::expm1l(3) | C++ Standard Libary | std::expm1,std::expm1f,std::expm1l(3) |
NAME¶
std::expm1,std::expm1f,std::expm1l - std::expm1,std::expm1f,std::expm1l
Synopsis¶
Defined in header <cmath>
float expm1 ( float arg ); (1) (since C++11)
float expm1f( float arg );
double expm1 ( double arg ); (2) (since C++11)
long double expm1 ( long double arg ); (3) (since C++11)
long double expm1l( long double arg );
double expm1 ( IntegralType arg ); (4) (since C++11)
1-3) Computes the e (Euler's number, 2.7182818) raised to the given power
arg, minus
1.0. This function is more accurate than the expression std::exp(arg)-1.0 if
arg is
close to zero.
4) A set of overloads or a function template accepting an argument of any
integral
type. Equivalent to (2) (the argument is cast to double).
Parameters¶
arg - value of floating-point or Integral type
Return value¶
If no errors occur earg
-1 is returned.
If a range error due to overflow occurs, +HUGE_VAL, +HUGE_VALF, or +HUGE_VALL
is
returned.
If a range error occurs due to underflow, the correct result (after rounding)
is
returned.
Error handling¶
Errors are reported as specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC
60559),
* If the argument is ±0, it is returned, unmodified
* If the argument is -∞, -1 is returned
* If the argument is +∞, +∞ is returned
* If the argument is NaN, NaN is returned
Notes¶
The functions std::expm1 and std::log1p are useful for financial
calculations, for
example, when calculating small daily interest rates: (1+x)n
-1 can be expressed as std::expm1(n * std::log1p(x)). These functions also
simplify
writing accurate inverse hyperbolic functions.
For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg
Example¶
// Run this code
#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";
}
Possible output:¶
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
See also¶
exp
expf returns e raised to the given power (\({\small e^x}\)e^x)
expl (function)
(C++11)
(C++11)
exp2
exp2f
exp2l returns 2 raised to the given power (\({\small 2^x}\)2^x)
(C++11) (function)
(C++11)
(C++11)
log1p
log1pf natural logarithm (to base e) of 1 plus the given number (\({\small
log1pl \ln{(1+x)} }\)ln(1+x))
(C++11) (function)
(C++11)
(C++11)
2022.07.31 | http://cppreference.com |