std::bit_width
From cppreference.com
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 + 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 base2 logarithm of x
, with any fractional part discarded.
Notes
This function is equivalent to return std::numeric_limits<T>::digits  std::countl_zero(x);.
Featuretest 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 behaviorchanging 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 
See also
(C++20) 
counts the number of consecutive 0 bits, starting from the most significant bit (function template) 