std::exp, std::expf, std::expl - cppreference.com (original) (raw)
Defined in header | ||
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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) | |
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 | ||
template< /*math-floating-point*/ V > constexpr /*deduced-simd-t*/<V> exp ( const V& v_num ); | (S) | (since C++26) |
Additional overloads (since C++11) | ||
Defined in header | ||
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) |
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[edit] Parameters
num | - | floating-point or integer value |
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[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
#include #include #include #include #include #include #include // #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:🔢: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
exp2exp2fexp2l(C++11)(C++11)(C++11) | returns 2 raised to the given power (\({\small 2^x}\)2x) (function) [edit] |
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expm1expm1fexpm1l(C++11)(C++11)(C++11) | returns e raised to the given power, minus 1 (\({\small e^x-1}\)ex-1) (function) [edit] |
loglogflogl(C++11)(C++11) | computes natural (base e) logarithm (\({\small\ln{x}}\)ln(x)) (function) [edit] |
exp(std::complex) | complex base e exponential (function template) [edit] |
exp(std::valarray) | applies the function std::exp to each element of valarray (function template) [edit] |
C documentation for exp |