std::bit_width

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(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

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);.

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';
    }
}

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

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

See also