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

Parameters

Return value

If no errors occur, the base- e exponential of num ( e num ) is returned.

If a range error occurs due to overflow, +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";
}

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

See also