std::bit_width
Defined in header <bit> |
||
---|---|---|
template< class T > constexpr T bit_width( T x ) noexcept; |
(since C++20) |
If x
is not zero, calculates the number of bits needed to store the value x
, that is, \(1 + \lfloor \log_2(x) \rfloor\)1 + floor(log
2(x)). If x
is zero, returns zero.
This overload participates in overload resolution only if T
is an unsigned integer type (that is, unsigned char
, unsigned short
, unsigned int
, unsigned long
, unsigned long long
, or an extended unsigned integer type).
Parameters
x | - | unsigned integer value |
Return value
Zero if x
is zero; otherwise, one plus the base-2 logarithm of x
, with any fractional part discarded.
Notes
This function is equivalent to return std::numeric_limits<T>::digits - std::countl_zero(x);
.
Feature testing macro: __cpp_lib_int_pow2
.
Example
#include <bit> #include <bitset> #include <iostream> int main() { for (unsigned x{0}; x != 8; ++x) { std::cout << "bit_width( " << std::bitset<4>{x} << " ) = " << std::bit_width(x) << '\n'; } }
Output:
bit_width( 0000 ) = 0 bit_width( 0001 ) = 1 bit_width( 0010 ) = 2 bit_width( 0011 ) = 2 bit_width( 0100 ) = 3 bit_width( 0101 ) = 3 bit_width( 0110 ) = 3 bit_width( 0111 ) = 3
See also
(C++20)
|
counts the number of consecutive 0 bits, starting from the most significant bit (function template) |
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