std::log1p, std::log1pf, std::log1pl
| Defined in header <cmath>
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| (1) | ||
float log1p ( float num ); double log1p ( double num ); |
(until C++23) | |
| /*floating-point-type*/ log1p ( /*floating-point-type*/ num ); |
(since C++23) (constexpr since C++26) |
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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) |
| SIMD overload (since C++26) |
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| Defined in header <simd>
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| template< /*math-floating-point*/ V > constexpr /*deduced-simd-t*/<V> |
(S) | (since C++26) |
| Additional overloads (since C++11) |
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| Defined in header <cmath>
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template< class Integer > double log1p ( Integer num ); |
(A) | (constexpr since C++26) |
std::log1p for all cv-unqualified floating-point types as the type of the parameter.(since C++23)|
S) The SIMD overload performs an element-wise
std::log1p on v_num.
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(since C++26) |
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A) Additional overloads are provided for all integer types, which are treated as double.
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(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 argument num of integer type, std::log1p(num) has the same effect as std::log1p(static_cast<double>(num)).
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 raisedSee also
| (C++11)(C++11) |
computes natural (base e) logarithm (ln(x)) (function) |
| (C++11)(C++11) |
computes common (base 10) logarithm (log10(x)) (function) |
| (C++11)(C++11)(C++11) |
base 2 logarithm of the given number (log2(x)) (function) |
| (C++11)(C++11)(C++11) |
returns e raised to the given power, minus 1 (ex-1) (function) |
| C documentation for log1p
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