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table of contents

std::log1p,std::log1pf,std::log1pl(3) | C++ Standard Libary | std::log1p,std::log1pf,std::log1pl(3) |

# NAME¶

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

# Synopsis¶

Defined in header <cmath>

float log1p ( float num );

double log1p ( double num ); (until C++23)

long double log1p ( long double num );

/* floating-point-type */ (since C++23)

log1p ( /* floating-point-type */ num ); (constexpr since C++26)

float log1pf( float num ); **(1)** **(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 > (A) (constexpr since C++26)

double log1p ( Integer num );

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)

A) Additional overloads are provided for all integer types, which are
*(since C++11)*

treated as double.

# 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 (until C++23)

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)).

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 (since
C++23)

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.

# Example¶

// Run this code

#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

# See also¶

log

logf computes natural (base e) logarithm (\({\small\ln{x}}\)ln(x))

logl *(function)*

*(C++11)*

*(C++11)*

log10

log10f computes common (base 10) logarithm
(\({\small\log_{10}{x}}\)log[10](x))

log10l *(function)*

*(C++11)*

*(C++11)*

log2

log2f

log2l base 2 logarithm of the given number (\({\small\log_{2}{x}}\)log[2](x))

*(C++11)* *(function)*

*(C++11)*

*(C++11)*

expm1

expm1f

expm1l returns e raised to the given power, minus one (\({\small
e^x-1}\)e^x-1)

*(C++11)* *(function)*

*(C++11)*

*(C++11)*

C documentation for

log1p

2024.06.10 | http://cppreference.com |