std::exp, std::expf, std::expl
From cppreference.com
| Defined in header <cmath>
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| (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) |
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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) |
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| Defined in header <simd>
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template< /*math-floating-point*/ V > constexpr /*deduced-simd-t*/<V> exp ( const V& v_num ); |
(S) | (since C++26) |
| Additional overloads (since C++11) |
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| Defined in header <cmath>
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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)|
S) The SIMD overload performs an element-wise
std::exp on v_num.
|
(since C++26) |
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A) Additional overloads are provided for all integer types, which are treated as
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 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
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, 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
See also
(C++11)(C++11)(C++11) |
returns 2 raised to the given power (2x) (function) |
(C++11)(C++11)(C++11) |
returns e raised to the given power, minus 1 (ex-1) (function) |
(C++11)(C++11) |
computes natural (base e) logarithm (ln(x)) (function) |
| complex base e exponential (function template) | |
| applies the function std::exp to each element of valarray (function template) | |
C documentation for exp
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