Exp

Defined in header <cmath>.

Description

Computes e (Euler's number, 2.7182818...) raised to the given power num.
The library provides overloads of std::exp for all cv-unqualified floating-point types as the type of the parameter num.

Declarations

C++23

// 1)
/* floating-point-type */ exp( /* floating-point-type */ num );
// 2)
float expf( float num  );
// 3)
long double expl( long double num );

Additional Overloads

// 4)
template< class Integer >
double exp ( Integer num );

C++11

// 1)
float exp ( float num );
// 2)
double exp ( double num );
// 3)
long double exp ( long double num );
// 4)
float expf( float num  );
// 5)
long double expl( long double num );

Additional Overloads

// 6)
template< class Integer >
double exp ( Integer num );

Parameters

​num - floating-point or integer value

Return value

If no errors occur, the base-e exponential of num 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 Additional Overloads. 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)).

Examples

#include <cerrno>
#include <cfenv>
#include <cmath>
#include <cstring>
#include <iostream>
#include <iomanip>
 
// #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 Result

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