template< class T >
constexpr int bit_width( T x ) noexcept;
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(log2(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).
Zero if x is zero; otherwise, one plus the base-2 logarithm of x, with any fractional part discarded.
[edit] NotesThis function is equivalent to return std::numeric_limits<T>::digits - std::countl_zero(x);.
[edit] Example#include <bit>
#include <bitset>
#include <iostream>
int main()
{
for (unsigned x{}; x != 010; ++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[edit] Defect reports
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR Applied to Behavior as published Correct behavior LWG 3656 C++20 the return type ofbit_width is the same as the type of its function argument made it int [edit] See also counts the number of consecutive â0â bits, starting from the most significant bit
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