double log2 ( double num );
/*floating-point-type*/
log2 ( /*floating-point-type*/ num );
float log2f( float num );
(2) (since C++11)long double log2l( long double num );
(3) (since C++11)constexpr /*deduced-simd-t*/<V>
template< class Integer >
double log2 ( Integer num );
Computes the
binary (base-2) logarithmof
num.
The library provides overloads ofstd::log2
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 treated as double.
(since C++11) [edit] Parameters num - floating-point or integer value [edit] Return valueIf no errors occur, the base-2 logarithm of num (log2(num) or lb(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.
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),
For integer num, the binary logarithm can be interpreted as the zero-based index of the most significant 1 bit in the input.
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::log2(num) has the same effect as std::log2(static_cast<double>(num)).
[edit] Example#include <cerrno> #include <cfenv> #include <cmath> #include <cstring> #include <iostream> // #pragma STDC FENV_ACCESS ON int main() { std::cout << "log2(65536) = " << std::log2(65536) << '\n' << "log2(0.125) = " << std::log2(0.125) << '\n' << "log2(0x020f) = " << std::log2(0x020f) << " (highest set bit is in position 9)\n" << "base-5 logarithm of 125 = " << std::log2(125) / std::log2(5) << '\n'; // special values std::cout << "log2(1) = " << std::log2(1) << '\n' << "log2(+Inf) = " << std::log2(INFINITY) << '\n'; // error handling errno = 0; std::feclearexcept(FE_ALL_EXCEPT); std::cout << "log2(0) = " << std::log2(0) << '\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:
log2(65536) = 16 log2(0.125) = -3 log2(0x020f) = 9.04166 (highest set bit is in position 9) base-5 logarithm of 125 = 3 log2(1) = 0 log2(+Inf) = inf log2(0) = -inf errno == ERANGE: Numerical result out of range FE_DIVBYZERO raised[edit] See also computes natural (base e) logarithm (\({\small\ln{x}}\)ln(x))
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