NAME

std::exp,std::expf,std::expl - std::exp,std::expf,std::expl

Synopsis

Defined in header <cmath>
float exp ( float arg );
float expf( float arg ); (since C++11)
double exp ( double arg ); (1) (2)
long double exp ( long double arg );
long double expl( long double arg ); (3) (since C++11)
double exp ( IntegralType arg ); (4) (since C++11)

1-3) Computes e (Euler's number, 2.7182818...) raised to the given power arg
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, the base-e exponential of arg (earg
) 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, 1 is returned
* If the argument is -∞, +0 is returned
* If the argument is +∞, +∞ is returned
* If the argument is NaN, NaN is returned

Notes

For IEEE-compatible type double, overflow is guaranteed if 709.8 < arg, and
underflow is guaranteed if arg < -708.4

Example

// Run this code

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cerrno>
#include <cstring>
#include <cfenv>
// #pragma STDC FENV_ACCESS ON
int main()
{
std::cout << "exp(1) = e¹ = " << std::setprecision(16) << std::exp(1) << '\n'
<< "FV of $100, continuously compounded at 3% for 1 year = "
<< std::setprecision(6) << 100*std::exp(0.03) << '\n';
// special values
std::cout << "exp(-0) = " << std::exp(-0.0) << '\n'
<< "exp(-Inf) = " << std::exp(-INFINITY) << '\n';
// error handling
errno = 0;
std::feclearexcept(FE_ALL_EXCEPT);
std::cout << "exp(710) = " << std::exp(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:

exp(1) = e¹ = 2.718281828459045
FV of $100, continuously compounded at 3% for 1 year = 103.045
exp(-0) = 1
exp(-Inf) = 0
exp(710) = inf
errno == ERANGE: Numerical result out of range
FE_OVERFLOW raised