Log1p

Defined in header <cmath>.

Description

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. Additional Overloads are provided for all integer types, which are treated as double.

Declarations

C++23

// 1)
/* floating-point-type */ log1p( /* floating-point-type */ num );
// 2)
float log1pf( float num );
// 3)
long double log1pl( long double num );

Additional Overloads

// 4)
template< class Integer >
double log1p ( Integer num );

C++11

// 1)
float log1p ( float num );
// 2)
double log1p ( double num );
// 3)
long double log1p ( long double num );
// 4)
float log1pf( float num );
// 5)
long double log1pl( long double num );

Additional Overloads

// 6)
template< class Integer >
double log1p ( Integer num );

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 zero.

Pole error may occur if num is zero.

If the implementation supports IEEE floating-point arithmetic (IEC 60559):

If the argument is ±0, -∞ is returned and FE_DIVBYZERO is raised
If the argument is 1, +0 is returned
If the argument is negative, 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 an 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 Additional Overloads. 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)). (do 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.

Examples

#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 Result

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