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

Defined in header `<cmath>`
	(1)
float exp ( float num ); double exp ( double num ); long double exp ( long double num );		(until C++23)
/* floating-point-type / exp ( / floating-point-type */ num );		(since C++23) (constexpr since C++26)
float expf( float num );	(2)	(since C++11) (constexpr since C++26)
long double expl( long double num );	(3)	(since C++11) (constexpr since C++26)
Additional overloads (since C++11)
Defined in header `<cmath>`
template< class Integer > double exp ( Integer num );	(A)	(constexpr since C++26)

2.7182818...

double

(since C++11)

Parameters

num

floating-point or integer value

Return value

If no errors occur, the base-e exponential of num (enum
) 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 < num, and underflow is guaranteed if num < -708.4.

The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their argument num of integer type, std::exp(num) has the same effect as std::exp(static_cast<double>(num)).

Example

#include <cerrno>
#include <cfenv>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <numbers>
 
// #pragma STDC FENV_ACCESS ON
 
consteval double approx_e()
{
    long double e {1.0};
    for (auto fac {1ull}, n{1llu}; n != 18; ++n, fac *= n)
        e += 1.0 / fac;
    return e;
}
 
int main()
{
    std::cout << std::setprecision(16)
              << "exp(1) = e¹ = " << std::exp(1) << '\n'
              << "numbers::e  = " << std::numbers::e << '\n'
              << "approx_e    = " << approx_e() << '\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
numbers::e  = 2.718281828459045
approx_e    = 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