A RetroSearch Logo

Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Search Query:

Showing content from https://en.cppreference.com/w/cpp/language/../keyword/../language/../numeric/math/exp.html below:

std::exp, std::expf, std::expl - cppreference.com

(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) template< /*math-floating-point*/ V >

constexpr /*deduced-simd-t*/<V>

            exp ( const V& v_num ); (S) (since C++26)

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

(A) (constexpr since C++26) 1-3)

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.(since C++23)

A) Additional overloads are provided for all integer types, which are treated as double.

(since C++11) [edit] Parameters num - floating-point or integer value [edit] Return value

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

[edit] Error handling

Errors are reported as specified in math_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

[edit] 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)).

[edit] 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
[edit] See also

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4