double log1p ( double num );
/*floating-point-type*/
log1p ( /*floating-point-type*/ num );
float log1pf( float num );
(2) (since C++11)long double log1pl( long double num );
(3) (since C++11)constexpr /*deduced-simd-t*/<V>
template< class Integer >
double log1p ( Integer num );
Computes the
natural (base-e) logarithmof
1 + num. This function is more precise than the expression
std::log(1 + num)if
numis close to zero.
The library provides overloads ofstd::log1p
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 valueIf no errors occur ln(1+num) is returned.
If a domain error occurs, an implementation-defined value is returned (NaN where supported).
If a pole error 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 handlingErrors are reported as specified in math_errhandling.
Domain error occurs if num is less than -1.
Pole error may occur if num is -1.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
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.
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::log1p(num) has the same effect as std::log1p(static_cast<double>(num)).
[edit] Example#include <cerrno> #include <cfenv> #include <cmath> #include <cstring> #include <iostream> // #pragma STDC FENV_ACCESS ON int main() { std::cout << "log1p(0) = " << log1p(0) << '\n' << "Interest earned in 2 days on $100, compounded daily at 1%\n" << " on a 30/360 calendar = " << 100 * expm1(2 * log1p(0.01 / 360)) << '\n' << "log(1+1e-16) = " << std::log(1 + 1e-16) << ", but log1p(1e-16) = " << std::log1p(1e-16) << '\n'; // special values std::cout << "log1p(-0) = " << std::log1p(-0.0) << '\n' << "log1p(+Inf) = " << std::log1p(INFINITY) << '\n'; // error handling errno = 0; std::feclearexcept(FE_ALL_EXCEPT); std::cout << "log1p(-1) = " << std::log1p(-1) << '\n'; if (errno == ERANGE) std::cout << " errno == ERANGE: " << std::strerror(errno) << '\n'; if (std::fetestexcept(FE_DIVBYZERO)) std::cout << " FE_DIVBYZERO raised\n"; }
Possible output:
log1p(0) = 0 Interest earned in 2 days on $100, compounded daily at 1% on a 30/360 calendar = 0.00555563 log(1+1e-16) = 0, but log1p(1e-16) = 1e-16 log1p(-0) = -0 log1p(+Inf) = inf log1p(-1) = -inf errno == ERANGE: Result too large FE_DIVBYZERO raised[edit] See also computes natural (base e) logarithm (\({\small\ln{x}}\)ln(x))
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