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

Defined in header `<cmath>`
	(1)
float log1p ( float num ); double log1p ( double num ); long double log1p ( long double num );		(until C++23)
/* floating-point-type / log1p ( / floating-point-type */ num );		(since C++23) (constexpr since C++26)
float log1pf( float num );	(2)	(since C++11) (constexpr since C++26)
long double log1pl( long double num );	(3)	(since C++11) (constexpr since C++26)
Additional overloads (since C++11)
Defined in header `<cmath>`
template< class Integer > double log1p ( Integer num );	(A)	(constexpr since C++26)

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)

double

(since C++11)

Parameters

num

floating-point or integer value

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.

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

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 first argument num1 and second argument num2:

If num1 or num2 has type long double, then std::log1p(num1, num2) has the same effect as std::log1p(static_cast<long double>(num1), static_cast<long double>(num2)).
Otherwise, if num1 and/or num2 has type double or an integer type, then std::log1p(num1, num2) has the same effect as std::log1p(static_cast<double>(num1), static_cast<double>(num2)).
Otherwise, if num1 or num2 has type float, then std::log1p(num1, num2) has the same effect as std::log1p(static_cast<float>(num1), static_cast<float>(num2)).

(until C++23)

If num1 and num2 have arithmetic types, then std::log1p(num1, num2) has the same effect as std::log1p(static_cast</* common-floating-point-type */>(num1), static_cast</* common-floating-point-type */>(num2)), where /* common-floating-point-type */ is the floating-point type with the greatest floating-point conversion rank and greatest floating-point conversion subrank between the types of num1 and num2, arguments of integer type are considered to have the same floating-point conversion rank as double.

If no such floating-point type with the greatest rank and subrank exists, then overload resolution does not result in a usable candidate from the overloads provided.

(since C++23)

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