std::expm1, std::expm1f, std::expm1l - cppreference.com (original) (raw)
| Defined in header | ||
|---|---|---|
| (1) | ||
| float expm1 ( float num ); double expm1 ( double num ); long double expm1 ( long double num ); | (until C++23) | |
| /*floating-point-type*/ expm1 ( /*floating-point-type*/ num ); | (since C++23) (constexpr since C++26) | |
| float expm1f( float num ); | (2) | (since C++11) (constexpr since C++26) |
| long double expm1l( 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> expm1 ( const V& v_num ); | (S) | (since C++26) |
| Additional overloads (since C++11) | ||
| Defined in header | ||
| template< class Integer > double expm1 ( Integer num ); | (A) | (constexpr since C++26) |
1-3) Computes the e (Euler's number, 2.7182818...) raised to the given power num, minus 1.0. This function is more accurate than the expression std::exp(num) - 1.0 if num is close to zero. The library provides overloads of std::expm1 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 enum
-1 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.
[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, it is returned, unmodified.
- If the argument is -∞, -1 is returned.
- If the argument is +∞, +∞ is returned.
- If the argument is NaN, NaN is returned.
[edit] Notes
The functions std::expm1 and std::log1p are useful for financial calculations, for example, when calculating small daily interest rates: (1+x)n
-1 can be expressed as std::expm1(n * std::log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.
For IEEE-compatible type double, overflow is guaranteed if 709.8 < num.
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::expm1(num) has the same effect as std::expm1(static_cast<double>(num)).
[edit] Example
#include #include #include #include #include // #pragma STDC FENV_ACCESS ON int main() { std::cout << "expm1(1) = " << std::expm1(1) << '\n' << "Interest earned in 2 days on $100, compounded daily at 1%\n" << " on a 30/360 calendar = " << 100 * std::expm1(2 * std::log1p(0.01 / 360)) << '\n' << "exp(1e-16)-1 = " << std::exp(1e-16) - 1 << ", but expm1(1e-16) = " << std::expm1(1e-16) << '\n'; // special values std::cout << "expm1(-0) = " << std::expm1(-0.0) << '\n' << "expm1(-Inf) = " << std::expm1(-INFINITY) << '\n'; // error handling errno = 0; std::feclearexcept(FE_ALL_EXCEPT); std::cout << "expm1(710) = " << std::expm1(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:
expm1(1) = 1.71828 Interest earned in 2 days on $100, compounded daily at 1% on a 30/360 calendar = 0.00555563 exp(1e-16)-1 = 0, but expm1(1e-16) = 1e-16 expm1(-0) = -0 expm1(-Inf) = -1 expm1(710) = inf errno == ERANGE: Result too large FE_OVERFLOW raised
[edit] See also
| | returns e raised to the given power (\({\small e^x}\)ex) (function) [edit] | | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | | returns 2 raised to the given power (\({\small 2^x}\)2x) (function) [edit] | | | natural logarithm (to base e) of 1 plus the given number (\({\small\ln{(1+x)}}\)ln(1+x)) (function) [edit] | | |