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std::log1p, std::log1pf, std::log1pl

Defined in header `<cmath>`
float log1p ( float arg ); float log1pf( float arg );	(1)	(since C++11)
double log1p ( double arg );	(2)	(since C++11)
long double log1p ( long double arg ); long double log1pl( long double arg );	(3)	(since C++11)
double log1p ( IntegralType arg );	(4)	(since C++11)

1-3) Computes the natural (base e) logarithm of 1+arg. This function is more precise than the expression std::log(1+arg) if arg is close to zero.

4) A set of overloads or a function template accepting an argument of any integral type. Equivalent to 2) (the argument is cast to double).

Parameters

arg

value of floating-point or Integral type

Return value

If no errors occur ln(1+arg) 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.

Error handling

Errors are reported as specified in math_errhandling.

Domain error occurs if arg is less than -1.

Pole error may occur if arg 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

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.

Example

#include <iostream>
#include <cfenv>
#include <cmath>
#include <cerrno>
#include <cstring>
#pragma STDC FENV_ACCESS ON
int main()
{
    std::cout << "log1p(0) = " << log1p(0) << '\n'
              << "Interest earned in 2 days on 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)
              << " 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 on $100, compounded daily at 1%
 on a 30/360 calendar = 0.00555563
log(1+1e-16) = 0 log1p(1e-16) = 1e-16
log1p(-0) = -0
log1p(+Inf) = inf
log1p(-1) = -inf
    errno == ERANGE: Result too large
    FE_DIVBYZERO raised