std::log1p, std::log1pf, std::log1pl - cppreference.com (original) (raw)

1-3) Computes the natural (base-e) logarithm of 1 + num. This function is more precise than the expression std::log(1 + num) if num is close to zero. The library provides overloads of std::log1p for all cv-unqualified floating-point types as the type of the parameter.(since C++23)

[edit] Parameters

[edit] Return value

If 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 handling

Errors 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),

• If the argument is ±0, it is returned unmodified.
• If the argument is -1, -∞ is returned and FE_DIVBYZERO is raised.
• If the argument is less than -1, NaN is returned and FE_INVALID is raised.
• 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.

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 #include #include #include #include // #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)) (function) [edit] | | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | | computes common (base 10) logarithm (\({\small\log_{10}{x}}\)log10(x)) (function) [edit] | | | base 2 logarithm of the given number (\({\small\log_{2}{x}}\)log2(x)) (function) [edit] | | | returns e raised to the given power, minus 1 (\({\small e^x-1}\)ex-1) (function) [edit] | | |