std::bit_width

Defined in header `<bit>`
template< class T > constexpr int 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-test macro	Value	Std	Comment
`__cpp_lib_int_pow2`	202002L	(C++20)	Integral power-of-2 operations

Example

Run this code

#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

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 of `bit_width` is the same as the type of its function argument	made it int

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