NAME

std::bit_width - std::bit_width

Synopsis

Defined in header <bit>
template< class T > (since C++20)
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

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: __cpp_lib_int_pow2

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 made it int
type of its function argument