std::exp, std::expf, std::expl
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Defined in header
<cmath>
(1)
float exp ( float num );
(until C++23)
double exp ( double num );
/*floating-point-type*/
exp ( /*floating-point-type*/ num );
(since C++23) exp ( /*floating-point-type*/ num );
(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)
SIMD overload (since C++26)
Defined in header
<simd>
template< /*math-floating-point*/ V >
(S)
(since C++26)
constexpr /*deduced-simd-t*/<V>
Additional overloads (since C++11)
Defined in header
<cmath>
template< class Integer >
double exp ( Integer num );
(A)
(constexpr since C++26)
double exp ( Integer num );
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)S) The SIMD overload performs an element-wise
std::exp on v_num.- (See math-floating-point and deduced-simd-t for their definitions.)
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),
- 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.
[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
Run this code
#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,ドル 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,ドル 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
(C++11)(C++11)(C++11)
(function) [edit]
C documentation for exp